C H A P T E R 2 Polynomial and Rational Functions

CHAPTER 2
Polynomial and Rational Functions
Section 2.1 Quadratic Functions and Models . . . . . . . . . . . . . 136
Section 2.2 Polynomial Functions of Higher Degree . . . . . . . . . 151
Section 2.3 Polynomial and Synthetic Division . . . . . . . . . . . . 168
Section 2.4 Complex Numbers . . . . . . . . . . . . . . . . . . . . 180
Section 2.5 Zeros of Polynomial Functions . . . . . . . . . . . . . . 187
Section 2.6 Rational Functions . . . . . . . . . . . . . . . . . . . . 205
Section 2.7 Nonlinear Inequalities . . . . . . . . . . . . . . . . . . 222
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

CHAPTER 2
Polynomial and Rational Functions
Section 2.1 Quadratic Functions and Models
136
You should know the following facts about parabolas.
■ f x ax is a quadratic function, and its graph is a parabola.
■ If a > 0, the parabola opens upward and the vertex is the point with the minimum y-value.
If a < 0, the parabola opens downward and the vertex is the point with the maximum y-value.
■ The vertex is b2a, f b2a.
■ To find the x-intercepts (if any), solve
2 bx c, a 0,
ax 2 bx c 0.
■ The standard form of the equation of a parabola is
f x ax h 2 k
where a 0.
(a) The vertex is h, k.
(b) The axis is the vertical line x h.
Vocabulary Check
1. nonnegative integer; real 2. quadratic; parabola 3. axis or axis of symmetry
4. positive; minimum 5. negative; maximum
1. f x x 2 opens upward and has vertex 2, 0.
Matches graph (g).
2 2. f x x 4 opens upward and has vertex 4, 0.
Matches graph (c).
2
3. f x x opens upward and has vertex 0, 2.
Matches graph (b).
2 2 4. f x 3 x opens downward and has vertex 0, 3.
Matches graph (h).
2
5. f x 4 x 2 opens downward
and has vertex 2, 4. Matches graph (f).
2 x 22 4 6. f x x 1 opens upward and has vertex
1, 2. Matches graph (a).
2 2
7. f x x 3 opens downward and has
vertex 3, 2. Matches graph (e).
2 2 8. f x x 4 opens downward and has vertex 4, 0.
Matches graph (d).
2

9. (a)
(c)
10. (a)
(c)
y 1
2 x2 (b)
Vertical shrink
y 3
2 x2
5
4
3
2
1
−3 −2 −1 1 2 3
−1
5
4
3
2
1
−3 −2 −1 1 2 3
−1
Vertical stretch
y x 2 1 (b)
5
4
3
2
−3 −2 −1 1 2 3
−1
Vertical translation one unit upward
y x 2 3
10
8
6
−6 −4 −2 2 4 6
−2
y
y
y
y
Vertical translation three units upward
x
x
x
x
Section 2.1 Quadratic Functions and Models 137
(d)
(d)
y 1
8 x2
Vertical shrink and reflection in the x-axis
y 3x 2
6
4
2
−6 −4 4 6
−2
−4
−6
6
4
2
−6 −4 −2 2 4 6
Vertical stretch and reflection in the x-axis
y x 2 1
4
3
2
1
−3 −2 2 3
−2
Vertical translation one unit downward
y x 2 3
8
6
4
−6 –4 4 6
−4
y
y
y
y
Vertical translation three units downward
x
x
x
x

138 Chapter 2 Polynomial and Rational Functions
11. (a)
(c)
12. (a)
(c)
y x 12 (b) y 3x2 1
−2 −1 1 2 3 4
−1
Horizontal translation one unit to the right
y 1
3 x2 3
−6
5
4
3
y
8
6
4
2
−4
y
−2 2 6
−2
x
x
Horizontal stretch and a vertical translation three units
downward
y 1
2x 2 2 1 (b)
Horizontal translation two units to the right, vertical
1
shrink each
y-value is multiplied by 2, reflection in
the x-axis, and vertical translation one unit upward
−8
−6
−4
8
6
4
y
−6 −4 −2 2 6 8 10
y 1
2x 2 2 1
6
4
2
−4
−6
−8
y
2 4 6
x
x
Horizontal translation two units to the left, vertical
1
shrink each
y-value is multiplied by 2, reflection in
x-axis, and vertical translation one unit downward
(d)
5
4
3
−3 −2 −1
−1
Horizontal shrink and a vertical translation one unit upward
y x 3 2
10
−8 −6 −4 −2 2 4
−2
y
8
2
1
y
2 3
Horizontal translation three units to the left
y 1
2x 1 2 3
10
x
x
Horizontal translation one unit to the right, horizontal
stretch (each x-value is multiplied by 2), and vertical
translation three units downward
(d) y 2x 12 4
8
6
4
−8 −6 −4 2 6 8
−4
−6
7
4
3
2
1
−4 −3 −2 −1 1 2 3 4
−1
y
y
x
x
Horizontal translation one unit to the left, horizontal
1
shrink each
x-value is multiplied by 2, and vertical translation
four units upward

13.
f x x 2 5 14.
Vertex:
Axis of symmetry: x 0 or the y-axis
Find x-intercepts:
x 2 5 0
x 2 5
0, 5
x ±5
x-intercepts:
5, 0, 5, 0
1
2 x2 4 0
x 2 8
x-intercepts:
x ±8 ±22
22, 0, 22, 0
2
1
−4 −3 −1 1 3 4
−2
−3
−6
y
3
2
1
−4 −3 −1 1 2 3 4
−2
−3
−5
x
x
Section 2.1 Quadratic Functions and Models 139
hx 25 x 2
Vertex:
0, 25
Axis of symmetry:
Find x-intercepts:
25 x 2 0
x 2 25
x ±5
x-intercepts: ±5, 0
16 1
4 x2 0
x 2 64
x ±8
x-intercepts: ±8, 0
x 0
15. f x
Vertex: 0, 4
Axis of symmetry: x 0 or the y-axis
Find x-intercepts:
y
1
2x2 4 1
2x 02 4 16.
Vertex: 0, 16
Axis of symmetry: x 0
Find x-intercepts:
17.
19.
f x x 5 y
2 6 18.
Vertex:
5, 6
Axis of symmetry:
Find x-intercepts:
x 5 2 6 0
x 5 2 6
x 5 ±6
x 5
x 5 ± 6
x-intercepts: 5 6, 0, 5 6, 0
20
16
12
−20 −12 4 8
−8
x
f x 16 1
4 x2 1
4 x2 16
f x x 6 2 3
Vertex:
6, 3
Axis of symmetry:
Find x-intercepts:
x 6 2 3 0
x 6 2 3
Not possible for real x
No x-intercepts
x 6
hx x2 8x 16 x 42 20. gx x2 2x 1 x 12 Vertex:
4, 0
Axis of symmetry: x 4
x-intercept: 4, 0
20
16
12
8
4
y
−4 4 8 12 16
x
Vertex:
1, 0
Axis of symmetry: x 1
x-intercept: 1, 0
30
−20 −10
10 20
y
18
12
9
6
3
y
−9 −6 −3
−3
50
40
30
20
10
y
3 6 9
−20 −10 10 20 30
−4 −3 −2 −1 1 2
6
5
4
3
2
1
y
x
x
x
x

140 Chapter 2 Polynomial and Rational Functions
21.
fx x 2 x 5
4
Vertex:
x2 x 1 1 5
4 4 4
1
x 2 2
1
1
, 1 2
Axis of symmetry:
Find x-intercepts:
x2 x 5
0
4
x
Not a real number
No x-intercepts
x 1
2
1 ± 1 5
2
5
4
3
1
y
−2 −1 1 2 3
23. f x x2 2x 5 24.
Vertex:
x 2 2x 1 1 5
x 1 2 6
Axis of symmetry:
Find x-intercepts:
x-intercepts: 1 6, 0, 1 6, 0
−4 2 6
−2
−4
1, 6
x 2 2x 5 0
x 2 2x 5 0
6
y
x
x 1
2 ± 4 20
2
1 ± 6
x
x
22.
f x x 2 3x 1
4
Vertex:
x2 3x 9 9 1
4 4 4
3
x 2 2
2
3, 2 2
Axis of symmetry:
Find x-intercepts:
x2 3x 1
0
4
x
x 3
2
3 ±9 1
2
3
± 2
2
x-intercepts: 3 ± 2, 0 2
−5 −4 −3 −2 −1 1 2
f x x 2 4x 1 x 2 4x 1
Vertex:
2, 5
Axis of symmetry:
Find x-intercepts:
x-intercepts:
2 ± 5, 0
−6 −5 −3 −2 −1 1 2
5
4
2
1
−2
−3
y
x
4
3
2
1
−2
−3
x 2 4x 4 4 1
x 2 2 5
x 2
x 2 4x 1 0
x 2 4x 1 0
x
4 ± 16 4
2
2 ± 5
y
x

25.
27.
29.
hx 4x 2 4x 21
Vertex:
Vertex:
4x2 x 1 1
4 4 21 4
1
4x 2 2
20
1
, 20 2
Axis of symmetry:
Find x-intercepts:
4x 2 4x 21 0
x
f x 1
4 x2 2x 12
1
4x 2 8x 16 1
416 12
1
4x 4 2 16
4, 16
Axis of symmetry:
Find x-intercepts:
1
4 x2 2x 12 0
x 2 8x 48 0
x 4x 12 0
x 4 or x 12
x 1
2
4 ± 16 336
24
Not a real number ⇒ No x-intercepts
x 4
x-intercepts: 4, 0, 12, 0
−8 −4
4 8
−8
4
−12
−16
−20
y
4 8 16
f x x 2 2x 3 x 1 2 4 30.
Vertex:
1, 4
Axis of symmetry: x 1
x-intercepts: 3, 0, 1, 0
20
10
y
−8 7
5
−5
x
x
Section 2.1 Quadratic Functions and Models 141
26.
28.
2x2 1
f x 2x
x 1
2 2 x 1
Vertex:
1
2x 4 2
2 1
1
16
1
2x 4 2
7
8
1 7
,
4 8
Axis of symmetry:
Find x- intercepts:
2 x 2 x 1 0
x
Not a real number
No x-intercepts
Vertex:
Axis of symmetry:
Find x-intercepts:
1
3 x2 3x 6 0
x 2 9x 18 0
x 3x 6 0
x 1
4
1 ± 1 8
22
f x 1
3x 2 3x 6
1
3x 2 9x 6
1
3x2 9x 81
4 1
381 4 6
1
3x 9
9 3
2 , 4
2 2
3
4
x 9
2
x-intercepts: 3, 0, 6, 0
f x x 2 x 30
x 2 x 30
6
5
4
3
1
−3 −2 −1 1 2 3
2
−2
−2
−4
−6
x2 x 1
4 1
4 30
x 1
2 2 121
4
Vertex:
Axis of symmetry: x
x-intercepts: 6, 0, 5, 0
1
2
y
y
4 6
8 10
1
2, 121
4 −10 10
35
−80
x
x

142 Chapter 2 Polynomial and Rational Functions
31.
33.
gx x 2 8x 11 x 4 2 5 32.
Vertex:
4, 5
Axis of symmetry:
x 4
x-intercepts: 4 ± 5, 0
Vertex:
2x 4 2 1
4, 1
Axis of symmetry:
x 4
x-intercepts: 4 ± 1
22, 0
14
−18 12
f x 2x 48
2 16x 31 34.
−6
−6 12
35. gx
Vertex: 2, 3
4
1
2x2 4x 2 1
2x 22 3 36.
Axis of symmetry:
x 2
x-intercepts: 2 ± 6, 0
−12
−8 4
−4
f x x 2 10x 14
Vertex:
x 2 10x 25 25 14
x 5 2 11
5, 11 −20 10
Axis of symmetry:
x 5
x-intercepts: 5 ± 11, 0
f x 4x 2 24x 41
Vertex:
4x 2 6x 41
4x 2 6x 9 36 41
4x 3 2 5
3, 5
Axis of symmetry:
No x-intercepts
Vertex:
Axis of symmetry:
x 3
f x 3
5x 2 6x 5
3
5x 2 6x 9 27
5 3
3
5x 32 42
5
x 3
x-intercepts: 3 ± 14, 0
37. 1, 0 is the vertex.
38. 0, 1 is the vertex.
y ax 1 2 0 ax 1 2
Since the graph passes through the point 0, 1, we have:
1 a0 1 2
1 a
y 1x 1 2 x 1 2
5
−15
0
0 6
3, 42
5 −14 10
f x ax 0 2 1 ax 2 1
−20
Since the graph passes through 1, 0,
0 a1 2 1
1 a.
So, y x 2 1.
39. 1, 4 is the vertex.
40. 2, 1
is the vertex.
y ax 1 2 4
Since the graph passes through the point 1, 0, we have:
0 a1 1 2 4
4 4a
1 a
y 1x 1 2 4 x 1 2 4
f x ax 2 2 1
Since the graph passes through 0, 3,
3 a0 2 2 1
3 4a 1
4 4a
1 a.
So, y x 2 2 1.
6
−10

41. 2, 2 is the vertex.
42. 2, 0 is the vertex.
y ax 2 2 2
Since the graph passes through the point 1, 0, we have:
0 a1 2 2 2
2 a
y 2x 2 2 2
Section 2.1 Quadratic Functions and Models 143
f x ax 2 2 0 ax 2 2
Since the graph passes through 3, 2,
2 a3 2 2
2 a.
So, y 2x 2 2 .
43. 2, 5 is the vertex.
44. 4, 1 is the vertex.
f x ax 2 2 5
Since the graph passes through the point 0, 9, we have:
9 a0 2 2 5
4 4a
1 a
fx 1x 2 2 5 x 2 2 5
f x ax 4 2 1
Since the graph passes through 2, 3,
3 a2 4 2 1
3 4a 1
4 4a
1 a.
So, f x x 4 2 1.
45. 3, 4 is the vertex.
46. 2, 3 is the vertex.
fx ax 3 2 4
Since the graph passes through the point 1, 2, we have:
2 a1 3 2 4
2 4a
1
2 a
fx 1
2x 3 2 4
f x ax 2 2 3
Since the graph passes through 0, 2,
2 a0 2 2 3
2 4a 3
1 4a
1
4 a.
So, f x 1
4x 2 2 3.
47. 5, 12 is the vertex.
48. 2, 2 is the vertex.
fx ax 5 2 12
Since the graph passes through the point 7, 15, we have:
15 a7 5 2 12
3 4a ⇒ a 3
4
f x 3
4x 5 2 12
f x ax 2 2 2
Since the graph passes through 1, 0,
0 a1 2 2 2
0 a 2
2 a.
So, f x 2x 2 2 2.
49. is the vertex.
1 3
4 , 2 50. 2 , 3 is the vertex.
f x ax 1
4 2 3
2
Since the graph passes through the point
we have:
3
2
0 a2 1
4 2 3
2
49
16a ⇒ a 24 49
fx 24
49x 1
4 2 3
2
2, 0,
5
f x ax 5
2 2
Since the graph passes through 2, 4,
4 a2 5
2 2
3
4
4 81 3
4 a 4
19
4
4
81
4 a
19
81 a.
3
4
f x 19
81x 5
2 2
3
4
So, .

144 Chapter 2 Polynomial and Rational Functions
51. , 0 is the vertex.
52. 6, 6 is the vertex.
53.
5
2
f x ax 5
2 2
Since the graph passes through the point
we have:
16
3
16
3
57. fx x 4
x-intercepts: 0, 0, (4,0
2 4x 58. f x 2x
x-intercepts: 0, 0, 5, 0
2 10x
0 x 2 4x
0 xx 4)
x 0
or
x 4
−4 8
The x-intercepts and the solutions of f x 0 are the same.
−4
0 2x 2 10x
0 2xx 5
2x 0 ⇒ x 0
x 5 0 ⇒ x 5
59. fx x 12
x-intercepts: 3, 0, 6, 0
2 9x 18 60.
x-intercepts:
0 x 2 9x 18
0 x 3)x 6
x 3
5
a7 2 2 2
a
f x 16
3 x 5
2 2
or
x 6
The x-intercepts and the solutions of f x 0 are the same.
−8
7
2 , 16 3 ,
y x 2 16 54.
x-intercepts:
0 x 2 16
x 2 16
x ±4
±4, 0
−4
16
f x ax 6 2 6
Since the graph passes through
3
2 a61 10 62 6
3 1
2 100a 6
9
2 1
100a
450 a.
So, f x 450x 6 . 2 6
y x 2 6x 9
x-intercept:
3, 0
0 x 2 6x 9
0 x 3 2
x 3 0 ⇒ x 3
55. y x
x-intercepts: 5, 0, 1, 0
2 4x 5 56. y 2x
x-intercepts:
2 5x 3
0 x 2 4x 5
0 x 5x 1
x 5
or x 1
1 2 , 0, 3, 0
0 2x 2 5x 3
0 2x 1x 3
2 x 1 0 ⇒ x 1
2
x 3 0 ⇒ x 3
−1 6
The x-intercepts and the solutions of f x 0 are the same.
f x x 2 8x 20
0 x 2 8x 20
0 x 2x 10
x 2 0 ⇒ x 2
x 10 0 ⇒ x 10
61 3
10 , 2,
2, 0, 10, 0 −4 12
The x-intercepts and the solutions of f x 0 are the same.
14
−6
10
−40

61.
fx 2x 10
2 7x 30 62.
x-intercepts:
0 2x 2 7x 30
5
2, 0, 6, 0
x or x 6
The x-intercepts and the solutions of f x 0 are the same.
5
0 2x 5)x 6
−40
2
fx 1
−5
10
Section 2.1 Quadratic Functions and Models 145
f x 4x 2 25x 21
x-intercepts:
7, 0, 3
4, 0
0 4x 2 25x 21
0 x 74x 3
x 7 0 ⇒ x 7
4x 3 0 ⇒ x 3
4
63.
2x 10
x-intercepts: 1, 0, 7, 0
2 6x 7 64.
x-intercepts: 15, 0, 3, 0
0 1
2x 2 6x 7
0 x 2 6x 7
0 x 1x 7
x 1
or
x 7
The x-intercepts and the solutions of f x 0 are the same.
x 1x 3
x 2 2x 3
gx x 1x 3
x 1x 3
x 2 2x 3
x 2 2x 3
−10
−6
opens downward
Note: f x ax 1x 3 has x-intercepts 1, 0
and 3, 0 for all real numbers a 0.
14
−9 2
The x-intercepts and the solutions of f x 0 are the same.
f x 7
10x 2 12x 45
0 7
10x 2 12x 45
0 x 15x 3
x 15 0 ⇒ x 15
x 3 0 ⇒ x 3
65. f x x 1x 3 opens upward
66. f x x 5x 5
x 2 10x
gx x 0x10
x 2 10x
opens downward
Note: f x ax 0x 10 axx 10 has
x-intercepts 0, 0 and 10, 0 for all real numbers
a 0.
The x-intercepts and the solutions of f x 0 are the same.
x 5x 5
x 2 25, opens upward
gx f x, opens downward
gx x 2 25
67. f x x 0x 10 opens upward
68. f x x 4x 8
f x x 3x 1
69. 22 opens upward
70.
x 3x 1
22
x 32x 1
2x 2 7x 3
gx 2x 2 7x 3
2x 2 7x 3
opens downward
Note: has x-intercepts
and for all real numbers a 0.
1
f x ax 32x 1
3, 0 2, 0
10
−70
10
−18 4
Note: f x ax has x-intercepts 5, 0 and
5, 0 for all real numbers a 0.
2 25
x 2 12x 32, opens upward
gx f x, opens downward
gx x 2 12x 32
Note: f x ax 4x 8 has x-intercepts 4, 0 and
8, 0 for all real numbers a 0.
f x 2x 5
2x 2
2x2 1
2x
2x 5
5
2x 2
2x 2 x 10, opens upward
gx f x, opens downward
gx 2x 2 x 10
5
f x ax
Note: 2x 2 has x-intercepts
and 2, 0 for all real numbers a 0.
−60
5
2 , 0

146 Chapter 2 Polynomial and Rational Functions
71. Let x the first number and y the second number. 72. Let x first number and y second number. Then,
Then the sum is
x y S, y S x. The product is
75. (a)
x y 110 ⇒ y 110 x.
The product is Px xy x110 x 110x x2 .
Px x 2 110x
x 2 110x 3025 3025
x 55 2 3025
x 55 2 3025
The maximum value of the product occurs at the vertex of
Px and is 3025. This happens when x y 55.
The maximum value of the product occurs at the vertex
of Px and is 72. This happens when x 12 and
y 24 122 6. Thus, the numbers are 12 and 6.
(c)
y
1
2 x2 24x 144 144
1
2 x 122 144 1
2 x 122 72
4x 3y 200 ⇒ y 1
4
200 4x 50 x
3 3
A 2xy 2x 4
8
8x50 x
50 x x50 x
3 3 3
2000
0
0
This area is maximum when feet and
y feet.
100
x 25
3 3313
—CONTINUED—
x
x
60
Px xy xS x.
Px Sx x 2
x 2 Sx
x2 Sx S2
4
S
x 2 2
S2
4
S2
4
The maximum value of the product occurs at the vertex
of and is S This happens when x y S2.
2 Px 4.
73. Let the first number and the second number.
Then the sum is
The product is
Px 1
2 x2 24 x
Px xy x 2
24x
.
x y
74. Let the first number and the second number.
24 x
x 2y 24 ⇒ y .
2
Then the sum is
The product is
Px 1
3 x2 42 x
Px xy x 3
42x
.
x
y
42 x
x 3y 42 ⇒ y .
3
1
3 x2 42x 441 441
1
3 x 212 441 1
3 x 212 147
The maximum value of the product occurs at the vertex
of Px and is 147. This happens when x 21 and
42 21
y 7. Thus, the numbers are 21 and 7.
3
(b) This area is maximum when
feet and y feet.
100
x A
x 25
3 3313
5 600
10 1066
15 1400
20 1600
25 1666
30 1600
2
3
2
3

75. —CONTINUED—
(d)
A 8
x50 x
3
8
3 x2 50x
8
3 x2 50x 625 625
8
3 x 252 625
8
3 x 252 5000
3
The maximum area occurs at the vertex and is
square feet. This happens when feet and
feet. The dimensions
are feet by 33 feet.
1
50003
x 25
y 200 4253 1003
2x 50
76. (a) Radius of semicircular ends of track:
Distance around two semicircular parts of track:
d 2r 2 1
r
y y 2 1
2 y
77.
(b) Distance traveled around track in one lap:
d y 2x 200
3
y 200 2x
y 4
9 x2 24
x 12
9
200 2x
y
Section 2.1 Quadratic Functions and Models 147
(c) Area of rectangular region:
200 2x
A xy x
2
2
2
The area is maximum when and
200 250
y 100
x 50
The vertex occurs at The maximum height is y3 feet.
4
9 32 24
b 249
3.
3 12 16
2a 249 9
78. y
(a) The ball height when it is punted is the y-intercept.
16
2025 x2 9
x 1.5
5
y 16
2025 02 9
0 1.5 1.5 feet
5
(b) The vertex occurs at x b
2a
95
2162025
The maximum height is
—CONTINUED—
f 3645 16
32
2025 3645
32 2
9
5 3645
32 1.5
6561
64
6561
32
3645
32 .
1.5 6561
64
(e) They are all identical.
feet and y 33 feet
1
x 25
3
1
200x 2x2
13,122
64
x2 100x
x2 100x 2500 2500
x 502 5000
96
64
.
6657
feet 104.02 feet.
64

148 Chapter 2 Polynomial and Rational Functions
78. —CONTINUED—
79.
The vertex occurs at x
The cost is minimum when x 20 fixtures.
b
C 800 10x 0.25x
10
20.
2a 20.25 2 0.25x2 10x 800 80.
81.
83.
85.
(c) The length of the punt is the positive x-intercept.
C 100,000 110x 0.045x 2
The vertex occurs at
x 110
1222.
20.045
The cost is minimum when x 1222 units.
The vertex occurs at x
The profit is maximum when x 350,000 units.
b
2a
P 0.0002x
140
350,000.
20.0002 2 140x 250,000 82.
The vertex occurs at x
Because x is in hundreds of dollars, 20 100 2000
dollars is the amount spent on advertising that gives
maximum profit.
b
P 230 20x 0.5x
20
20.
2a 20.5 2
Rp 25p 2 1200p 84.
(a)
0 16
2025 x 2 9
x 1.5
5
x 95 ± 952 41.5162025
322025
x 0.83031 or x 228.64
The punt is approximately 228.64 ft.
R20 $14,000 thousand
R25 $14,375 thousand
R30 $13,500 thousand
(b) The revenue is a maximum at the vertex.
b 1200
24
2a 225
R24 14,400
The unit price that will yield a maximum revenue of
$14,400 thousand is $24.
C 4299 1.8t 1.36t 2 , 0 ≤ t ≤ 43
(a)
(c)
5000
0
0
43
C40 2051
209,128,0942051
Annually: 8879 cigarettes
48,308,590
8879
Daily: 24 cigarettes
366
1.8 ± 1.81312
0.01580247
R p 12p 2 150p
(a)
R$4 12$4 2 150$4 $408
R$6 12$6 2 150$6 $468
R$8 12$8 2 150$8 $432
(b) The vertex occurs at
p b 150
$6.25
2a 212
Revenue is maximum when price $6.25 per pet.
The maximum revenue is
f$6.25 12$6.25 2 150$6.25 $468.75.
(b) Vertex 0, 4299
The vertex occurs when y 4299 which is the
maximum average annual consumption. The warnings
may not have had an immediate effect, but over time
they and other findings about the health risks and the
increased cost of cigarettes have had an effect.

86. (a) and (c)
(b)
(d) 1996
(e) Vertex occurs at
(f)
950
4
650
88. (a) and (c)
y 4.303x 2 49.948x 886.28
Minimum occurs at year 1996.
x 18
31
12
x b 49.948
5.8
2a 24.303
y 4.30318 2 49.94818 886.28 1381.388
There will be approximately 1,381,000 hairdressers
and cosmetologists in 2008.
10 80
20
(b)
Section 2.1 Quadratic Functions and Models 149
87. (a)
(b)
0
25
−5
0.002s 2 0.005s 0.029 10
a 2, b 5, c 10,029
s 5 ± 52 4210,029
22
s
y 0.0082x 2 0.746x 13.47
2s 2 5s 10,029 0
5 ± 80,257
4
s 72.1, 69.6
100
2s 2 5s 29 10,000
The maximum speed if power is not to exceed
10 horsepower is 69.6 miles per hour.
(d) The maximum of the graph is at x 45.5, or about 45.5 mi/h.
Algebraically, the maximum occurs at
x mi/h.
b 0.746
45.5
2a 20.0082
89. True. The equation 12x has no real solution,
so the graph has no x-intercepts.
2 1 0 90. True. The vertex of is and the vertex of
is 5
f x 4 gx
4 , 71 4 .
91.
ax2 b
f x ax
x c
a 2 bx c
ax2 b b2
x
a 4a
b f 2a
b2 b2
c
4a 2a
2 b2
b
ax
2a 2
b2
c
4a
ax b
2a 2 4ac b2
4a
b2 b
a4a2 b c
2a
b2 2b 2 4ac
4a
4a2 c
b 4ac b2
So, the vertex occurs at ,
2a
4ac b2
4a
4a
b b
, f 2a 2a .
5 53
, 4

150 Chapter 2 Polynomial and Rational Functions
92. Conditions (a) and (d) are preferable because profits
would be increasing.
f x ax 2 bx c
94. If has two real zeros, then by the Quadratic Formula they are
x b ± b2 4ac
.
2a
The average of the zeros of f is
b b 2 4ac
2a
b b2 4ac
2a
2
This is the x-coordinate
of the vertex of the graph.
2b
2a
2 b
2a .
95. 4, 3 and 2, 1
96.
m
1 3 2
2 4 6 1
3
y 1 1
x 2
3
y 1 1 2
x
3 3
y 1 5
x
3 3
7 3
, 2 , m
2 2
y 2 3 7
x 2 2
y 2 3 21
x
2 4
y 3 13
x
2 4
97. and
The slope of the perpendicular line through is
and the y-intercept is
y 5
m b 3.
4x 3
5
m
0, 3
4
4
4x 5y 10 ⇒ y 5
4
5x 2 98. y 3x 2
m 3
For a parallel line, m 3. So, for 8, 4, the line is
y 4 3x 8
For Exercises 99–104, let and gx 8x2 f x 14x 3,
.
99.
101.
143 3 83 2 27
93. Yes. A graph of a quadratic equation whose vertex is
has only one x-intercept.
0, 0
y 4 3x 24
y 3x 20.
f g3 f 3 g3 100. g f 2 82 2 142 3 32 28 3 7
fg4 7 f 4 7g4 7
144 7 384 7 2
11 128
49 1408
49
102. f 141.5 3
1.5 g 81.52 24
18 4
3
103. f g1 fg1 f 8 148 3 109 104. g f 0 g f 0 g140 3 g3
105. Answers will vary.
83 2 72

Section 2.2 Polynomial Functions of Higher Degree
Section 2.2 Polynomial Functions of Higher Degree 151
You should know the following basic principles about polynomials.
■ is a polynomial function of degree n.
■ If f is of odd degree and
(a) then (b) then
1. 1.
2. 2.
■ If f is of even degree and
(a) then (b) then
1. 1.
2. 2.
■ The following are equivalent for a polynomial function.
(a) is a zero of a function.
(b) is a solution of the polynomial equation
(c) is a factor of the polynomial.
(d) is an x-intercept of the graph of f.
■ A polynomial of degree n has at most n distinct zeros and at most turning points.
■ A factor x a yields a repeated zero of x a of multiplicity k.
(a) If k is odd, the graph crosses the x-axis at x a.
(b) If k is even, the graph just touches the x-axis at x a.
■ If f is a polynomial function such that a < b and fa fb, then f takes on every value between fa and fb in the
interval a, b.
■ If you can find a value where a polynomial is positive and another value where it is negative, then there is at least one real
zero between the values.
k fx anx an > 0,
an < 0,
fx → as x → .
fx → as x → .
fx → as x → .
fx → as x → .
an > 0,
an < 0,
fx → as x → .
fx → as x → .
fx → as x → .
fx → as x → .
x a
x a
fx 0.
x a
a, 0
n 1
, k > 1,
n an1xn1 . . . a2x2 a1x a0 , an 0,
Vocabulary Check
1. continuous 2. Leading Coefficient Test 3.
4. solution; x a; x-intercept
5. touches; crosses 6. standard
7. Intermediate Value
n; n 1
1. fx 2x 3 is a line with y-intercept 0, 3.
2. fx x is a parabola with intercepts 0, 0 and
Matches graph (c).
4, 0 and opens upward. Matches graph (g).
2 4x
3. is a parabola with x-intercepts
0, 0 and and opens downward. Matches
graph (h).
5
fx 2x
2 , 0
2 5x 4. has intercepts
and Matches graph (f).
1 1
2 23, 0.
1
fx 2x 0, 1, 1, 0,
1
2 23, 0
3 3x 1
5. fx has intercepts 0, 0 and ±23, 0.
Matches graph (a).
1
4x4 3x2 6. fx has y-intercept
Matches graph (e).
1
3x3 x2 4
3
0, 4
3.
7. fx x has intercepts 0, 0 and 2, 0.
Matches graph (d).
4 2x3 8. fx has intercepts 0, 0, 1, 0,
1, 0, 3, 0, 3, 0. Matches graph (b).
1
5x5 2x3 9
5x

152 Chapter 2 Polynomial and Rational Functions
9.
(a) (b) fx x3 fx x 2 2
3
y x3 10.
y x 5
(a)
(c)
4
3
2
1
−3 −2 2 3 4 5
−2
−3
−4
fx x 1 5
−4 −3 1 2 3 4
−3
−4
Horizontal shift one unit to the left
f x 1 1
2 x5
y
Horizontal shift two units to the right Vertical shift two units downward
(c) (d) fx x 23 fx 2
1
−4 −3 −2 2 3 4
−2
−3
−4
2 x3
4
3
2
1
4
3
2
1
4
3
2
−4 −3 −2 2 3 4
−3
−4
y
y
y
x
x
Reflection in the x-axis and a vertical shrink Horizontal shift two units to the right and
a vertical shift two units downward
x
x
Reflection in the x-axis, vertical shrink each y-value
1
is multiplied by 2, and vertical shift one unit upward
(b)
(d)
3
2
1
−3 −2 1 2 4 5
−2
−3
−4
−5
fx x 5 1
y
3
2
1
−4 −3 −2 2 3 4
−4
−5
4
3
2
−4 −3 −2 1 2 3 4
−3
−4
Vertical shift one unit upward
f x 1
2x 1 5
−5 −4 −3 −2 1 2 3
y
y
4
3
2
1
−3
−4
y
x
x
x
x
Reflection in the x-axis, vertical shrink each y-value is
1
multiplied by 2, and horizontal shift one unit to the left

11.
(a) (b) fx x4 fx x 3 3
4
y x4 12.
Section 2.2 Polynomial Functions of Higher Degree 153
Horizontal shift three units to the left Vertical shift three units downward
(c) (d) fx 1
2x 14 fx 4 x4 −4 −3 −2 1 2 3 4
−1
Reflection in the x-axis and then a vertical
shift four units upward
Horizontal shift one unit to the right and a vertical shrink
each y-value is multiplied by
(e) (f) fx 1 2x 4 fx 2x 2
4 1
2
1
−4 −3 −2 −1
−1
−2
Vertical shift one unit upward and a horizontal shrink Vertical shift two units downward and a horizontal stretch
each y-value is multiplied by each y-value is multiplied by 1
1
y x 6
(a)
−5 −4 −3 −2 −1 1 2 3
6
5
3
2
1
−2
y
6
5
y
6
5
4
3
2
1
−2
fx 1
8 x 6
y
1
2 3 4
x
x
x
Vertical shrink each y-value is multiplied by and
reflection in the x-axis
—CONTINUED—
4
3
2
1
−4 −3 −2 2 3 4
−1
−2
−3
−4
y
x
2
1
8
−4 −3 −2 2 3 4
−4 −3 −2 −1 1 2 3 4
−4 −3 −1 1
−1
4
3
2
1
−4
−2
6
5
4
3
2
1
y
y
y
3 4
(b) fx x 2 6 4
−5 −4 −2 1 2 3
−4
y
x
x
x
x
Horizontal shift two units to the left and vertical shift four
units downward
2

154 Chapter 2 Polynomial and Rational Functions
12. —CONTINUED—
(c)
(e)
fx x 6 4 (d)
−4 −3 −2 2 3 4
Vertical shift four units downward
fx 1
4 x6
2
4
3
2
1
−8 −6 −2 2 6 8
−4
y
y
x
x
Horizontal stretch (each x-value is multiplied by 4),
and vertical shift two units downward
fx 1
4 x 6 1
4
3
2
−4 −3 −2 2 3 4
−1
−2
−3
−4
y
Reflection in the x-axis, vertical shrink each y-value is
multiplied by and vertical shift one unit upward
(f) fx 2x6 1
4,
1
y
−4 −3 −2 −1
−1
−2
1 2 3 4
x
x
Horizontal shrink each x-value is multiplied by 2, and
vertical shift one unit downward
13. fx
Degree: 3
1
Leading coefficient: 3
The degree is odd and the leading coefficient is positive.
The graph falls to the left and rises to the right.
1
3x3 5x 14. fx 2x
Degree: 2
Leading coefficient: 2
The degree is even and the leading coefficient is positive.
The graph rises to the left and rises to the right.
2 3x 1
15. gx 5
Degree: 2
Leading coefficient: 3
The degree is even and the leading coefficient is negative.
The graph falls to the left and falls to the right.
7
2x 3x2 16. hx 1 x
Degree: 6
Leading coefficient: 1
The degree is even and the leading coefficient is negative.
The graph falls to the left and falls to the right.
6
17. fx 2.1x
Degree: 5
Leading coefficient: 2.1
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
5 4x3 2 18. fx 2x
Degree: 5
Leading coefficient: 2
The degree is odd and the leading coefficient is positive.
The graph falls to the left and rises to the right.
5 5x 7.5
19.
fx 6 2x 4x 2 5x 3 20.
Degree: 3
Leading coefficient: 5
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
fx
Degree: 4
3
Leading coefficient: 4
The degree is even and the leading coefficient is positive.
The graph rises to the left and rises to the right.
3x4 2x 5
4
1

Section 2.2 Polynomial Functions of Higher Degree 155
21. ht
Degree: 2
2
3t2 5t 3 22. fs
Degree: 3
7
8 s3 5s2 7s 1
23.
25.
27.
29.
Leading coefficient:
2
3
The degree is even and the leading coefficient is
negative. The graph falls to the left and falls to the right.
fx 3x 3 9x 1; gx 3x 3 24.
8
−4 4
−8
−8
12
−20
g
g
f
f
fx x 4 4x 3 16x; gx x 4 26.
8
fx x 2 25 28. (a)
(a)
Zeros:
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c)
(a)
0 x 2 25 x 5x 5
Turning point: 1 (the vertex of the parabola)
−30
Zero:
x ±5
10
−30
(b) t 3 has a multiplicity of 2 (even multiplicity).
(c)
30
0 t 2 6t 9 t 3 2
Turning point: 1 (the vertex of the parabola)
−18
t 3
4
−20
18
Leading coefficient:
7
8
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
fx 1
3 x 3 3x 2, gx 1
3 x 3
f
g
−9 9
6
−6
fx 3x 4 6x 2 , gx 3x 4
f
g
−6 6
5
−3
x ±7, both with multiplicity 1
(b) Multiplicity of x 7is
1.
(c)
fx 49 x 2
Multiplicity of x 7 is 1.
There is one turning point.
−30
0 7 x7 x
ht t 2 6t 9 30. (a) fx x2 10x 25
55
−5
30
x 5, with multiplicity 2
(b) The multiplicity of x 5 is 2.
(c)
There is one turning point.
−25
0 x 5 2
25
−5
15

156 Chapter 2 Polynomial and Rational Functions
31.
fx 1
(a)
32. (a)
33.
34. (a)
35.
3x2 1
3
0 1
3 x 2 1 2
3 x 3
1
3 x 2 x 2
1
3 x 2x 1
Zeros:
x 2
3
x 2, x 1
(b) Each zero has a multiplicity of 1 (odd multiplicity).
Turning point: 1 (the vertex of the parabola)
fx 1
2 x2 5 3
x
2 2
1
a
2 ,
x 5 2 ± 5
5 37 ± 2 4
b 5
2 ,
5 ± 37
,
2
ft t 3 4t 2 4t
c 3
2
f x 3x 3 12x 2 3x
2 2 41 2 3
1
2
both with multiplicity 1
(a) 0 3x3 12x 2 3x 3xx2 4x 1
Zeros: x 0, x 2 ± 3 (by the Quadratic
Formula)
(b) Each zero has a multiplicity of 1 (odd multiplicity).
Turning points: 2
gx 5xx 2 2x 1 (b) The multiplicity of x 0 is 1.
0 5xx 2 2x 1
0 xx 2 2x 1
For x a 1, b 2, c 1.
2 2x 1,
x 2 ± 2 2 411
21
2 ± 8
2
1 ± 2
The zeros are 0, 1 2, and 1 2, all with
multiplicity 1.
(a) 0 t
Zeros: t 0, t 2
(b) t 0 has a multiplicity of 1 (odd multiplicity).
t 2 has a multiplicity of 2 (even multiplicity).
Turning points: 2
3 4t2 4t tt2 4t 4 tt 22 (c)
(c)
(c)
−6
−6
8
−24
The multiplicity of x 1 2 is 1.
The multiplicity of x 1 2 is 1.
There are two turning points.
−1
12
−16
4
−4
5 37
(b) The multiplicity of is 1.
2
(c)
(c)
5 37
The multiplicity of is 1.
2
There is one turning point.
−8
−7
5
−5
3
−5
6
4
6
3
8

36. (a)
0 with multiplicity 2, 4 and 5 with multiplicity 1.
(b) The multiplicity of x 0 is 2.
(c)
38. (a)
fx x 4 x 3 20x 2 37.
x 0, 4, 5
The multiplicity of x 5 is 1.
The multiplicity of x 4 is 1.
There are three turning points.
−6
0 x 2 x 2 x 20
0 x 2 x 4x 5
25
−150
x 0, ±2, all with multiplicity 1
(b) The multiplicity of x 0 is 1.
(c)
40. (a)
6
f x x 5 x 3 6x 39.
The multiplicity of x 2 is 1.
The multiplicity of x 2 is 1.
There are two turning points.
−9
0 xx 4 x 2 6
0 xx 2 3x 2 2
6
−6
x ±5, both with multiplicity 1
(b) The multiplicity of x 5 is 1.
(c)
The multiplicity of x 5 is 1.
There is one turning point.
−6
20
−60
9
6
Section 2.2 Polynomial Functions of Higher Degree 157
gt t 5 6t 3 9t
(a)
Zeros:
(b) t 0 has a multiplicity of 1 (odd multiplicity).
(c)
(a)
0 t 5 6t 3 9t tt 4 6t 2 9 tt 2 3 2
t ±3 each have a multiplicity of 2
(even multiplicity).
Turning points: 4
−9
tt 3 2
t 3 2
t 0, t ±3
6
−6
fx 5x 4 15x 2 10
No real zeros
(b) Turning point: 1
(c)
0 5x 4 15x 2 10
−4
5x 4 3x 2 2
5x 2 1x 2 2
f x 2x4 2x2 40 41. gx x3 3x2 4x 12
0 2x 4 2x 2 40
0 2x 2 4x 5x 5
(a)
Zeros:
40
−5
9
4
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c)
0 x 3 3x 2 4x 12 x 2 x 3 4x 3
Turning points: 2
−8
x 2 4x 3 x 2x 2x 3
x ± 2, x 3
4
−16
7

158 Chapter 2 Polynomial and Rational Functions
42. (a)
44.
x ±5, 4, all with multiplicity 1
(b) The multiplicity of x 5 is 1.
(c)
f x x 3 4x 2 25x 100 43.
The multiplicity of x 5 is 1.
The multiplicity of x 4 is 1.
There are two turning points.
−9
0 x 2 x 4 25x 4
0 x 2 25x 4
0 x 5x 5x 4
140
−20
9
y 4x 3 4x 2 8x 8 45.
(a)
(b)
(c)
−3
−11
(1, 0, 1.414214, 0, 1.414214, 0
0 4x 3 4x 2 8x 8
0 4x 2 x 1 8x 1
0 4x 2 8x 1
0 4x 2 2x 1
x ±2, 1
(d) The intercepts match part (b).
46. y
(a)
12
1
4 x 3x 2 9
47.
2
−18 18
−12
(b) 0, 0, 3, 0, 3, 0
fx x 2 10x
Note: fx ax 0x 10 axx 10
has zeros 0 and 10 for all real numbers a 0.
3
(c)
y 4x 3 20x 2 25x
(a)
(b) x-intercepts:
(c)
−2
12
−4
0 4x 3 20x 2 25x
0 x2x 5 2
x 0 or x 5
2
0, 0, 5 2 , 0
(d) The solutions are the same as the x-coordinates
of the x-intercepts.
y x 5 5x 3 4x
(a)
(b) x-intercepts: 0, 0, ±1, 0, ±2, 0
(c)
0
x 0, ±3
1
4x3x 2 9
−6 6
−4
0 x 5 5x 3 4x
0 xx 2 1x 2 4
4
0 xx 1x 1x 2x 2
x 0, ±1, ±2
x-intercepts: 0, 0, ±3, 0
(d) The intercepts match part (b).
(d) The solutions are the same as the x-coordinates
of the x-intercepts.
fx x 0x 10 48. f x x 0x 3
xx 3
x 2 3x
6
Note: f x axx 3 has zeros 0 and 3 for all real
numbers a.

49.
51.
53.
55.
57.
59.
61.
fx x 2x 6 50.
fx x 2x 6
fx x 2 4x 12
Note: fx ax 2x 6 has zeros 2 and 6
for all real numbers a 0.
fx x 0x 2x 3 52.
xx 2x 3
x 3 5x 2 6x
Note: fx axx 2x 3 has zeros 0, 2, 3 for
all real numbers a 0.
fx x 4x 3x 3x 0 54.
x 4x 2 9x
x 4 4x 3 9x 2 36x
Note: fx ax has these zeros
for all real numbers a 0.
4 4x3 9x2 36x
fx x 1 3 x 1 3
x 1 3 x 1 3
x 1 2 3 2
x 2 2x 1 3
x 2 2x 2
Note: fx ax has these zeros for all real
numbers a 0.
2 2x 2
fx x 2x 2
x 2 2 x 2 4x 4
Note: f x ax a 0, has degree 2 and
zero x 2.
2 4x 4,
f x x 3x 0x 1
xx 3x 1 x 3 2x 2 3x
Note: f x ax a 0, has degree 3 and
zeros x 3, 0, 1.
3 2x2 3x,
xx 3x 3 x 3 3x
Note: f x ax a 0, has degree 3 and zeros
x 0, 3, 3.
3 3x,
Section 2.2 Polynomial Functions of Higher Degree 159
56.
58.
60.
f x x 4x 5
x 4x 5
x 2 x 20
Note: f x ax 4x 5 has zeros 4 and 5 for all
real numbers a.
f x x 0x 2x 5
xx 2x 5
xx 2 7x 10
x 3 7x 2 10x
Note: f x axx 2x 5 has zeros 0, 2, 5 for all
real numbers a.
f x x 2x 1x 0x 1x 2
xx 2x 1x 1x 2
xx 2 4x 2 1
xx 4 5x 2 4
x 5 5x 3 4x
Note: f x axx 2x 1x 1x 2 has
zeros 2, 1, 0, 1, 2 for all real numbers a.
x 2x 42 f x x 2x 4 5 x 4 5
x 2x 4 5x 4 5
5
xx 4 2 5x 2x 4 2 10
x 3 8x 2 16x 5x 2x 2 16x 32 10
x 3 10x 2 27x 22
Note: f x ax has these zeros
for all real numbers a.
3 10x 2 27x 22
f x x 8x 4
x 8x 4 x 2 12x 32
Note: f x ax has degree 2
and zeros x 8 and 4.
2 12x 32, a 0,
f x x 2x 4x 7
x 2x 2 11x 28 x 3 9x 2 6x 56
Note: f x ax has degree
3 and zeros x 2, 4, and 7.
3 9x2 6x 56, a 0,
f x x 0x 3x 3 62. f x x 93 x3 27x2 243x 729
Note: f x ax has
degree 3 and zero x 9.
3 27x2 243x 729, a 0,

160 Chapter 2 Polynomial and Rational Functions
63.
64.
65.
66.
67.
f x x 5 2 x 1x 2 x 4 7x 3 3x 2 55x 50
or f x x 5x 12x 2 x4 x3 15x2 23x 10
or f x x 5x 1x 22 x4 17x2 36x 20
Note: Any nonzero scalar multiple of these functions would also have degree 4 and zeros x 5, 1, 2.
f x x 4x 1x 3x 6 x 4 4x 3 23x 2 54x 72
Note: f x ax has degree 4 and zeros x 4, 1, 3, and 6.
4 4x3 23x2 54x 72, a 0,
f x x 4 x 4 x 5 4x 4
or f x x3x 42 x5 8x4 16x3 or f x x2x 43 x5 12x4 48x3 64x2 or f x xx 44 x5 16x4 96x3 256x2 256x
Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x 0 and 4.
f x x 3 2 x 1x 5x 6 x 5 6x 4 22x 3 108x 2 189x 270
or
f x x 3x 1 2 x 5x 6 x 5 10x 4 14x 3 88x 2 183x 90
or f x x 3x 1x 52x 6 x5 14x 4 50x3 68x2 555x 450
or f x x 3x 1x 5x 62 x5 15x 4 59x3 63x2 648x 540
Note: Any nonzero multiple of these functions would also have degree 5 and zeros x 3, 1, 5, and 6.
f x x 3 9x xx 2 9 xx 3x 3
(a) Falls to the left; rises to the right
(b) Zeros: 0, 3, 3
(c)
x 3 2 1 0 1 2 3
(d)
f x
(−3, 0)
−12 −8 −4
−4
−8
0 10 8 0 8 10 0
12
4
y
4
(0, 0)
(3, 0)
8 12
69. f t
(a) Rises to the left; rises to the right
1
4t2 2t 15 1
4t 12 7
2
(b) No real zero (no x-intercepts)
(c)
t 1
0 1 2 3
f t
4.5
3.75
3.5
x
3.75
4.5
68.
gx x 4 4x 2 x 2 x 2x 2
(a) Rises to the left; rises to the right
(b) Zeros:
(c)
(d)
2, 0, 2
x 0.5 1 1.5 2.5
gx
0.94 3 3.94 14.1
4
3
(−2, 0)
2
1
(0, 0) (2, 0)
−4 −3 −1 1 3 4
−4
(d) The graph is a parabola with vertex
8
6
2
−4 −2 2 4
y
y
t
x
1, 7
2.

70.
72.
74.
gx x 2 10x 16 x 2x 8
(a) Falls to the left; falls to the right
(b) Zeros: 2, 8
(c)
(d)
10
8
6
4
2
x 1 3 5 7 9
gx
y
f x 1 x 3
7 5 9 5 7
(2, 0) (8, 0)
4 6 10
(a) Rises to the left; falls to the right
(b) Zero: 1
(c)
(d)
x 2 1 0 1 2
f x
9 2 1 0 7
3
2
−2 −1 2
−1
y
(1, 0)
f x 4x 3 4x 2 15x
(a) Rises to the left; falls to the right
(b) Zeros:
(c)
(d)
x4x 2 4x 15
x2x 52x 3
3 ( − , 0( 2
−4 −3 −2
3 5
2 , 0, 2
x
x 3 2 1 0 1 2 3
f x
x
99 18 7 0 15 14 27
20
16
12
8
4
y
(0, 0)
1
2
(
5
, 0( 2
3
4
x
Section 2.2 Polynomial Functions of Higher Degree 161
71.
73.
75.
f x x 3 3x 2 x 2 x 3
(a) Falls to the left; rises to the right
(b) Zeros: 0, 3
(c)
(d)
x 1 0 1 2 3
f x
1
−1 1 2 4
−3
−4
y
4
(0, 0) (3, 0)
0 2 4 0
fx 3x 3 15x 2 18x 3xx 2x 3
(a) Falls to the left; rises to the right
(b) Zeros: 0, 2, 3
(c)
(d)
x 0 1 2 2.5 3
f x
y
7
6
5
4
3
2
1
(0, 0) (2, 0) (3, 0)
−3 −2 −1
−1
1 4 5 6
−2
x
0 6 0 1.875 0 7.875
f x 5x 2 x 3 x 2 5 x
(a) Rises to the left; falls to the right
(b) Zeros:
(c)
(d)
−15
5
(−5, 0) (0, 0)
−10
0, 5
−20
y
5 10
x
x
3.5
x 5
4 3 2 1 0 1
f x
0 16 18 12 4 0 6

162 Chapter 2 Polynomial and Rational Functions
76.
77.
f x 48x 2 3x 4
(a) Rises to the left; rises to the right (d)
(b) Zeros:
(c)
(a) Falls to the left; rises to the right
(b) Zeros:
(c)
(d)
79. gt
(a) Falls to the left; falls to the right
1
4t 22t 22 (b) Zeros: 2, 2
(c)
(d)
3x 2 x 2 16
f x x 2 x 4
(−2, 0)
2
0, 4
x 1 0 1 2 3 4 5
f x
0, ±4
x 5 4 3 2 1 0 1 2 3 4 5
f x
675 0 189 144 45 0 45 144 189 0 675
y
5
(0, 0) (4, 0)
−4 −2 2 6 8
t 0 1 2 3
0 0 25
9
4
9
3 2 1
gt
25
4
−3 −1 1 2 3
−1
−2
−5
−6
y
0 3 8 9 0 25
(2, 0)
4
x
t
4
4
78. hx
(a) Falls to the left; rises to the right
1
3x3x 42 (b) Zeros: 0, 4
(c)
(d)
x 1 0 1 2 3 4 5
hx
14
12
10
8
6
4
y
25
3
(0, 0) (4, 0)
0 3 9 0
−4 −2 2 4 6 8 10 12
x
32
3
80. gx
(a) Falls to the left; rises to the right
1
10x 12x 33 (b) Zeros:
(c)
(d)
1, 3
6
4
y
2
(−1, 0) (3, 0)
−6 −4 −2 4 6 8
x
125
3
x 2 1 0 1 2 4
g x
(−4, 0)
−6
−2
100
−200
−300
y
(0, 0)
2
(4, 0)
12.5 0 2.7 3.2 0.9 2.5
6
x

81.
83.
85.
Zeros: 0, 2, 2 all of multiplicity 1
gx 1
5x 1 2 x 32x 9
−12
Zeros: 1 of multiplicity 2; 3 of multiplicity 1;
of multiplicity 1
9
2
fx x 3 4x xx 2x 2 82.
−9 9
14
−6
6
−6
fx x 3 3x 2 3
−5
10
−10
The function has three zeros.
They are in the intervals
1, 0, 1, 2 and 2, 3. They
are x 0.879, 1.347, 2.532.
87. gx 3x4 4x3 3
−5
10
−10
18
The function has two zeros.
They are in the intervals
2, 1 and 0, 1. They
are x 1.585, 0.779.
5
5
x
3
2
1
0 3
1 1
2
3 3
4 19
x
4
3
2
1
0
y 1
51
17
1
1
y 1
509
132
13
4
3
1 4
2 77
3 348
Section 2.2 Polynomial Functions of Higher Degree 163
84.
86.
88.
f x 1
4 x 4 2x 2
−9 9
6
−6
Zeros: 2.828 and 2.828 with multiplicity 1; 0, with
multiplicity 2
hx 1
5x 2 2 3x 5 2
21
−12 12
−3
Zeros: 2, both with multiplicity 2
5
3,
f x 0.11x 3 2.07x 2 9.81x 6.88
The function has three zeros. They are in the intervals
0, 1, 6, 7, and 11, 12. They are approximately 0.845,
6.385, and 11.588.
x y x y
−4
10
−10
The function has four zeros. They are
in the intervals 4, 3, 1, 0, 0, 1,
and 3, 4. They are approximately
±3.113 and ±0.556.
−4
10
−30
16
hx x 4 10x 2 3
4
0
1
6.88
0.97
2 5.34
3 6.89
4 6.28
5 4.17
6 1.12
7
8
9
10
11
1.91
4.56
6.07
5.78
3.03
12 2.84
x
4
3
2
1
99
0 3
1
2
3
y
6
21
6
6
21
6
4 99

164 Chapter 2 Polynomial and Rational Functions
89. (a)
90. (a)
(b)
x > 0,
12 x > 0,
x < 12
Domain: 0 < x < 6
212 x 46 xx
8x12 x6 x
6 x > 0
x < 6
(c)
Volume l w h 24 2x24 4xx (c)
91. (a) A l w 12 2xx 2x square inches
2 12x
(b)
Thus, Vx 36 2x36 2xx x36 2x2 .
(b) Domain:
(d)
Volume l w h
height x
length width 36 2x
The length and width must be positive.
3600
0 18
0
The maximum point on the graph occurs at x 6.
This agrees with the maximum found in part (c).
16 feet 192 inches
V l w h
0 < x < 18
12 2xx192
384x cubic inches
2 2304x
(c) Since x and 12 2x cannot be negative, we have
0 < x < 6 inches for the domain.
(d) When the volume is a
maximum with V 3456 in. . The
dimensions of the gutter cross-section
are 3 inches 6 inches 3 inches.
3
x V
x 3,
0 0
1 1920
2 3072
3 3456
4 3072
5 1920
6 0
(e)
The volume is a maximum of 3456 cubic inches when the
height is 6 inches and the length and width are each
24 inches. So the dimensions are 6 24 24 inches.
720
600
480
360
240
120
Box Box Box
Height Width Volume, V
1
2
3
4
5
6
7 736 272 636 26
36 27
3388
2 36 26
3456
V
36 21
36 22
36 23
36 24
36 25
1 2 3 4 5 6
136 21 2 1156
236 22 2 2048
336 23 2 2700
436 24 2 3136
536 25 2 3380
x 2.6 corresponds to a maximum of about
665 cubic inches.
4000
0
0
Maximum: 3, 3456
The maximum value is the same.
6
x
(f) No. The volume is a product of the constant length and
the cross-sectional area. The value of x would remain the
same; only the value of V would change if the length was
changed.

92. (a)
93.
(c)
200
7
140
V 4
3r 3 r 2 4r (b)
V 4
3r 3 4r 3
150
0
0
16
3 r 3
The model is a good fit to the actual data.
13
2
(d)
r ≥ 0
r 1.93 ft
Section 2.2 Polynomial Functions of Higher Degree 165
V 120 ft 3 16
3 r 3
length 4r 7.72 ft
y 1 0.139t 3 4.42t 2 51.1t 39 94.
95. Midwest:
97.
South:
y 218 $223.472 thousand $223,472
Since the models are both cubic functions with positive
leading coefficients, both will increase without bound
as t increases, thus should only be used for short term
projections.
(a)
y 0.056t 3 1.73t 2 23.8t 29
180
7
120
13
The data fit the model closely.
y 118 $259.368 thousand $259,368 96. Answers will vary.
G 0.003t 3 0.137t 2 0.458t 0.839,
60
−10 45
−5
(b) The tree is growing most rapidly at t 15.
102. (a) Degree: 3
Leading coefficient: Positive
(b) Degree: 2
Leading coefficient: Positive
2 ≤ t ≤ 34
(c)
Example: The median price of homes in the South are
all lower than those in the Midwest. The curves do
not intersect.
y 0.009t 2 0.274t 0.458
b 0.274
15.222
2a 20.009
y15.222 2.543
Vertex 15.22, 2.54
(d) The x-value of the vertex in part (c) is approximately
equal to the value found in part (b).
98. R
The point of diminishing returns (where the graph
changes from curving upward to curving downward)
occurs when x 200. The point is 200, 160 which
corresponds to spending $2,000,000 on advertising to
obtain a revenue of $160 million.
1
100,000x 3 600x2 99. False. A fifth degree polynomial can have at most four
turning points.
100. True. f x x 1 has one repeated solution.
6
(c) Degree: 4
Leading coefficient: Positive
(d) Degree: 5
Leading coefficient: Positive
101. True. A polynomial of degree 7 with a negative leading
coefficient rises to the left and falls to the right.

166 Chapter 2 Polynomial and Rational Functions
103. fx x is even.
4 ; fx
(f)
Even
gx 1
2
Vertical shrink
Even
(h) gx f f x f f x f x 4 x 44 x16 Even
1 4 f x 2x 104. (a) is decreasing.
y2 is increasing.
3
5x 25 y1
3
1
3x 25 1 (c)
Since is not always increasing or always decreasing,
cannot be written in the form ax h5 Hx x
Hx
Hx
k.
5 3x3 2x 1
8
−12 12
y
2
y1 −8
(b) The graph is either always increasing or always
decreasing. The behavior is determined by a. If
a > 0, gx will always be increasing. If a < 0,
gx will always be decreasing.
−9 9
105. 5x2 7x 24 5x 8x 3 106. 6x 3 61x2 10x x6x2 61x 10
107.
109.
(a)
gx fx 2
Vertical shift two units upward
gx f x 2
f x 2
gx
Even
(b) (c) gx f x x4 x 4
gx f x 2
Horizontal shift two units to the left Reflection in the y- axis. The graph looks the same.
4 x4 7x3 15x2 x24x2 7x 15 108. y3 216 y3 63 x 2 4x 5x 3
5
4
3
2
1
−3 −2 −1 1 2 3
−1
(d) gx fx x
Reflection in the x-axis
4
Neither odd nor even Even
(e) gx f
Horizontal stretch
1 1 4
2x 16x 2 x 2 x 28 0 110.
2x 7x 4 0
2 x 7 0 ⇒ x 7
2
x 4 0 ⇒ x 4
y
x
Even
(g) gx f x x ≥ 0
34 x344 x3 ,
Neither odd nor even
6
−6
x6x 1x 10
y 6y 2 6y 36
3x 2 22x 16 0
3x 2x 8 0
3x 2 0
x 2
3
or
x 8 0
or x 8

111.
113.
115.
117.
12x 2 11x 5 0 112.
3x 14x 5 0
3x 1 0 ⇒ x 1
3
4x 5 0 ⇒ x 5
4
x 2 2x 21 0 114.
x 2 2x 1 2 21 1 0
x 1 2 22 0
x 1 2 22
x 1 ±22
x 1 ± 22
2 x 2 5x 20 0 116.
2 x2 5 x 20 0
2
5 2 x 4 2
185
2
0
8 x2 5
2 x 5
4 2 25
20 0
8
fx x 4 2
Common function:
5
x 4 2
185
16
y x 2
Transformation: Horizontal
shift four units to the left
x 5
±185
4 4
x
5 ± 185
4
−7 −6 −5 −4 −3 −2 −1 −1
7
6
5
4
3
2
1
y
1
x
Section 2.2 Polynomial Functions of Higher Degree 167
118.
x 2 24x 144 0
x 12 2 0
x 12 0
x 2 8x 2 0
x 4 2 14
x 12
x 2 8x 2
x 2 8x 16 2 16
x 4 ±14
3x 2 4x 9 0
x2 4
x 3 0
3
x2 4
x 3
3
x 4 ± 14
x2 4 4 4
x 3
3 9 9
2
x 3 2
31
9
x 2 31
± 3 9
x 2 31
±
3 3
x
f x 3 x 2
2 ± 31
3
Reflection in the x-axis
and vertical shift of three
units upward of y x 2
119. fx x 1 5 y
120. f x 7 x 6
Common function: y x
1
Horizontal shift of six units
Transformation: Horizontal
shift one unit to the left and
a vertical shift five units
downward
−3 −2 −1
−1
−2
−3
1 2 3
x
to the right, reflection in the
x-axis, and vertical shift of
seven units upward
of y x
−5
4
2
1
−4 −3 −1
−1
15
12
9
6
3
y
−2
−3
−4
y
1 3 4
−3 3 6 9 12 15
−3
x
x

168 Chapter 2 Polynomial and Rational Functions
121. fx 2x 9 y
122. f x 10
Common function: y x
Transformation: Vertical
stretch each
y-value is
multiplied by 2 ,
then a
vertical shift nine units
upward
−6
6
5
4
3
2
1
−3 −2 −1 1 2
x
Horizontal shift of three units
to the left, vertical shrink
each
y-value is multiplied
1
by 3, reflection in the x-axis
and vertical shift of ten
units upward of y x
1
3 x 3
Section 2.3 Polynomial and Synthetic Division
You should know the following basic techniques and principles of polynomial division.
■ The Division Algorithm (Long Division of Polynomials)
■ Synthetic Division
■ f k is equal to the remainder of fx divided by x k (the Remainder Theorem).
■ fk 0 if and only if x k is a factor of fx.
Vocabulary Check
1. f x is the dividend; dx is the divisor; gx is the quotient; rx is the remainder
y 1 x2
1. and
x 2
x 2 ) x 2 0x 0
x 2 2x
x 2
2x 0
2x 4
x
Thus, and y1 y2 .
2
4
x 2
x 2 x 2
3. y1 and
(a) and (b)
x5 3x3 x2 1
−9 9
6
−6
y 2 x 2 4
x 2
4
y 2 x 3 4x 4x
x 2 1
−2
9
8
7
6
5
4
3
2
1
y
−1 1 2
2. improper; proper 3. synthetic division 4. factor 5. remainder
(c)
3 4 5 6 7 8 9
2. and
39
x
Thus, and y1 y2 .
4 3x2 1
x2 x
5
2 8 39
x2 8x
5
2 8x
40
2 x
1
4 5x2 x2 5 ) x4 3x2 x
1
2 y2 x
8
2 8 39
x2 y1
5
x4 3x2 1
x2 5
x5 0x4 x3 x2 0x 1 ) x5 0x4 3x3 0x2 x
0x 0
3 4x
4x 3 0x 2 0x
4x 3 0x 2 4x
4x 0
x
Thus, and y1 y2. 5 3x3 x2 1 x3 4x 4x
x2 1
x

2x 4
4. and y2 x 3
x
(a) and (b)
2 y1
x 1
x3 2x2 5
x2 x 1
5.
7.
9.
8
−12 12
−8
x 3 ) 2x 2 10x 12
2x 2 6x
2x 2 10x 12
x 3
4x 12
4x 12
2x 4
4x 5 ) 4x 3 7x 2 11x 5
4x 3 5x 2
4x 3 7x 2 11x 5
4x 5
x 4 2x 3
0
(c)
2x 4 6.
12x 2 11x
12x 2 15x
x 4 5x 3 6x 2 x 2
x 2
x 2 3x 1 8.
4x 5
4x 5
0
x 2 3x 1
x 3 3x 2 1 10.
x 2 ) x 4 5x 3 6x 2 x 2
3x 3 6x 2
3x 3 6x 2
x 2
x 2
0
x 3 3x 2 1
Section 2.3 Polynomial and Synthetic Division 169
x 2 x 1 ) x 3 2x 2 0x 5
x 3 x 2 x
x 3
3x 2 x 5
3x 2 3x 3
2x 8
x
Thus, and y1 y2 .
3 2x2 5
x2 2x 4
x 3
x 1 x2 x 1
x 4 ) 5x 2 17x 12
5x 2 20x
5x 2 17x 12
x 4
5x 3
3x 12
3x 12
6x 3 4x 2
0
5x 3
3x 2 ) 6x 3 16x 2 17x 6
12x 2 17x
12x 2 8x
6x 3 16x 2 17x 6
3x 2
x 3 3x 2
x 3 4x 2 3x 12
x 3
2 x 2 4x 3
9x 6
9x 6
x 2 7x 18
x 3 ) x 3 4x 2 3x 12
7x 2 3x
7x 2 21x
0
2x 2 4x 3
18x 12
18x 54
42
x 2 7x 18 42
x 3

170 Chapter 2 Polynomial and Rational Functions
11.
13.
15.
16.
18.
x 2 ) 7x 3
7x 3
x 2
7x 14
11
7 11
x 2
2x 2 0x 1 ) 6x 3 10x 2 x 8
6x3 10x 2 x 8
2x 2 1
7 12.
6x 3 0x 2 3x
10x 2 2x 8
10x 2 0x 5
2x 3
3x 5
x 4 2x 3 3x 2
3x 5
2x 3
2x 2 1
x 2 2x 4
x 2 2x 3 ) x 4 0x 3 3x 2 0x 1
2x 3 0x 2 0x
2x 3 4x 2 6x
4x 2 6x 1
4x 2 8x 12
2x 11
⇒
x 3 0x 2 0x 1 ) x 5 0x 4 0x 3 0x 2 0x 7
x5 7
x3 1 x2 x2 7
x3 1
x 5 0x 4 0x 3 x 2
2 x
x 2 2x 1 ) 2x 3 4x 2 15x 5
2x 3 4x 2 2x
17x 5
2x 3 4x 2 15x 5
x 12 17x 5
2x
x 2 2x 1
14.
2 x 1 ) 8x 5
8x 4
9
8x 5
4
2x 1 9
2x 1
4
x 2 0x1 ) x 3 0x 2 0x 9
x 3 0x 2 x
x3 9
x2 x 9
x
1 x2 1
x4 3x2 1
x2 2x 3 x2 2x 11
2x 4
x2 2x 3
x 2 17.
x 2 7
19.
5 3
3
17
15
2
15
10
5
3x 3 17x 2 15x 25
x 5
x
x 9
x 3 3x 2 3x 1 ) x 4 0x 3 0x 2 0x 0
x 4 3x 3 3x 2 x
x4 x 13 x 3 6x2 8x 3
x 13 3x 3 3x 2 x 0
3x 3 9x 2 9x 3
25
25
0
6x 2 8x 3
3x 2 2x 5
x 3

20.
22.
24.
26.
28.
29.
31.
33.
3 5
5
18
15
3
7
9
2
5x 3 18x 2 7x 6
x 3
2 9
9
18
18
0
16
0
16
32
32
9x 3 18x 2 16x 32
x 2
6 3
3
16
18
2
3x 3 16x 2 72
x 6
2 5
5
0
10
10
5x 3 6x 8
x 2
3 1
0
12
12
6
6
0
5x 2 3x 2
0
72
72
0
9x 2 16
3x 2 2x 12
6
20
26
x 5 13x 4 120x 80
x 3
8 1
1
1
8
52
44
5x 2 10x 26 44
x 2
13
3
16
0
48
48
x 3 512
x 8 x2 8x 64
2 3
3
0
8
8
0
64
64
0
144
144
120
432
312
80
936
856
Section 2.3 Polynomial and Synthetic Division 171
21.
23.
25.
27.
2 4
4
8
8
0
9
0
9
4x 3 8x 2 9x 18
x 2
10 1
1
0
10
10
x 3 75x 250
x 10
4 5
x 4 16x 3 48x 2 144x 312 856
x 3
512
512
0
3x4 x 2 3x3 6x2 12x 24 48
x 2
6 1
1
180x x 4
0
6
6
0
6
6
0
12
12
0
36
36
0
24
24
180
216
36
0
48
48
0
216
216
x 6 x3 6x 2 36x 36 216
x 6
30.
32.
34.
5
6
20
14
5x 3 6x 2 8
x 4
6 10
10
50
60
10
0
56
56
10x 4 50x 3 800
x 6
9 1
1
0
9
9
0
81
81
18
18
0
4x 2 9
75
100
25
250
250
0
x 2 10x 25
8
224
232
5x 2 14x 56 232
x 4
0
60
60
0
360
360
x 3 729
x 9 x2 9x 81
2 3
3
0
6
6
0
12
12
800
2160
1360
10x 3 10x 2 60x 360 1360
x 6
729
729
0
0
24
24
0
48
48
3x 4
x 2 3x 3 6x 2 12x 24 48
x 2
1 1
1
2
1
3
5 3x 2x 2 x 3
x 1
3
3
6
5
6
11
x 2 3x 6 11
x 1

172 Chapter 2 Polynomial and Rational Functions
35.
37.
39.
41.
43.
1
2
4
4
16
2
14
23
7
30
4x 3 16x 2 23x 15
x 12
15
15
0
4x 2 14x 30
fx x 3 x 2 14x 11, k 4
4 1
1
1
4
3
14
12
2
11
8
3
fx x 4x 2 3x 2 3
f4 4 3 4 2 144 11 3
f x 15x 4 10x 3 6x 2 14, k 2
3
2
3
15
15
10
10
0
6
0
6
f 2
3 15 2
3 4 10 2
3 3 6 2
3 2 14 34
f x x
3
2
315x3 6x 4 34
3
0
4
4
14
8
3
34
3
f x x 3 3x 2 2x 14, k 2
2 1
1
3
2
3 2
2
2 32
32
14
6
8
f x x 2x 2 3 2x 32 8
f 2 2 3 32 2 22 14 8
f x 4x 3 6x 2 12x 4, k 1 3
1 3 4
4
6
4 43
2 43
12
10 23
2 23
36.
38.
40.
3
2
3
3
4
9
2
1
2
3x 3 4x 2 5
0
3
4
3
4
5
9
8
49
8
x 32 3x 2 1
2
2 1
f x x 1 34x 2 2 43x 2 23 0
f 1 3 41 3 3 61 3 2 121 3 4 0
44. f x 3x
2 2 3
8
10
6 32 2 42
3 8x2 10x 8, k 2 2
3
2 32
8 42
4
4
8
8
0
0
42.
1
5
2
7
11
14
3
x 3
4
f x x 3 5x 2 11x 8, k 2
8
6
2
f x x 2x 2 7x 3 2
8 20 22 8 2
49
8x 12
f 2 2 3 52 2 112 8
f x 10x 3 22x 2 3x 4, k 1
5
1
5
10
10
22
2
20
3
4
7
4
7
5
f 1 5 101 5 3 221 5 2 31 f x x
5 4
1
510x2 20x 7 13
5
13
5
2 22 3 65 13
25 25 5 4 25 5
f x x 3 2x 2 5x 4, k 5
5 1
1
2
5
2 5
5
25 5
25
4
10
f x x 5 x 2 2 5 x 25 6
f 5 5 3 25 2 55 4
55 10 55 4 6
f x x 2 2 3x 2 2 32 x 8 42 0
f 2 2 32 2 3 82 2 2 102 2 8
320 142 86 42 102 2 8
60 422 48 322 20 102 8
0
6

45. fx 4x
(a) 1 4 0 13
4 4
4 4 9
3 13x 10
(b) 2 4
(c)
(d)
(b)
(c)
1
2
f 1 1
f 2 4
4
4
f 1 2 4
8 4
4
4
0
2
2
0
32
32
f 8 1954
1
3
3
3
2 3
3
(d) 5 3
3
5
1
6
h1 3 5
3
h2 17
0
8
8
13
1
12
5
6
1
5
15
10
h5 199
13
256
243
10
2
8
13
16
3
10
9
1
10
6
4
10
2
8
10
50
40
10
6
4
10
1944
1954
1
8
3
5
3
1
16
17
1
200
199
Section 2.3 Polynomial and Synthetic Division 173
46. gx x
(a) 2 1 0 4 0
2 4 0
1 2 0 0
6 4x 4 3x 2 2
(b)
(c) 3 1
(d)
g2 14
4 1
g4 3122
1
g3 434
1 1
1
1
g1 2
0
3
3
0
1
1
0
4
4
4
9
5
4
1
3
4
16
12
0
15
15
0
3
3
3
0
3
0
48
48
47. hx 3x
(a) 3 3 5 10 1
9 42 96
3 14 32 97
h3 97
3 5x2 10x 1 48. f x 0.4x
(a) 1 0.4 1.6 0.7 0 2
0.4 1.2 0.5 0.5
0.4 1.2 0.5 0.5 2.5
f 1 2.5
4 1.6x 3 0.7x 2 2
49. 2 1
1
0
2
2
7
4
3
Zeros: 2, 3, 1
6
6
0
x 3 7x 6 x 2x 2 2x 3
x 2x 3x 1
(b) 2 0.4
(c)
f 2 20
5 0.4
0.4
0.4
1.6
2.0
0.4
f 5 65.5
(d) 10 0.4
50. 4 1
1.6
0.8
2.4
0.4
f 10 5668
1
0
4
4
Zeros: 4, 2, 6
0.7
2.0
2.7
1.6
4.0
5.6
28
16
12
0.7
4.8
5.5
0
13.5
13.5
0.7
56.0
56.7
48
48
0
0
11
11
3
3
0
0
6
6
3
45
48
2
67.5
65.5
0
567
567
2
22
20
x 3 28x 48 x 4x 2 4x 12
x 4x 6x 2
3
192
195
0
144
144
0
0
0
2
12
14
2
5670
5668
0
780
780
2
0
2
2
432
434
2
3120
3122

174 Chapter 2 Polynomial and Rational Functions
51.
53.
1
2
2
2
3 1
1
3 1
15
1
14
x 1
2 2x2 2x
14x 20
3 15x2 27x 10
2x 1x 2x 5
Zeros: 1
2 , 2, 5
1
x 3 2x 2 3x 6 x 3 x 3 x 2
Zeros: ±3, 2
55. 1 3 1
1
1 3 1
1
2 3
x
x 1x 1 3 x 1 3
3 3x2 2 x 1 3 x 1 3 x 1
Zeros: 1, 1 ± 3
56. 2 5 1
1 2
2
2
3
2 3
1
2 5 1
1
3
2
1
27
7
20
3
2
3
1 3
2 3
2 3
3
3 23
1 3
1
1
2 5
1 5
1 5
2 5
10
10
0
23
23
23
Zeros: 2 5, 2 5, 3
0
6
6
0
0
1 3
1 3
1 3
1 3
57. f x 2x Factors:
(a) 2 2 1 5 2
4 6 2
2 3 1 0
3 x2 5x 2;
3
1
1
0
0
13
7 35
6 35
6 35
6 35
2
2
0
x x 2 5x 2 5x 3
3 x 2 13x 3
Both are factors of f x since the remainders are zero.
0
3
3
0
x 2, x 1
52.
2
3
48
48
48x 3 80x 2 41x 6 x 2
348x 2 48x 9
Zeros: 2
3, 3
4, 1
4
54. 2 1
1
2 1
80
32
48
41
32
9
2
2
2 2
2 2
2
x x 2x 2x 2
3 2x2 1 2 0
2x 4
Zeros: 2, 2, 2
6
6
0
x 2
34x 312x 3
3x 24x 34x 1
2
22 2
22
22
22
4
4
(b) The remaining factor of f x
is 2x 1.
(c) f x 2x 1x 2x 1
(d) Zeros:
1
2, 2, 1
(e)
7
−6 6
−1
0

58. f x 3x Factors: x 3, x 2
3 2x2 19x 6;
(a)
3 3
3
2 3
3
2
9
7
7
6
1
2
2
19
21
(b) The remaining factor is 3x 1.
(b)
4 1
1
1
4
3
0
2
10
12
2
6
6
59. f x x Factors:
(a) 5 1 4 15 58 40
5 5 50 40
1 1 10 8 0
4 4x3 15x2 58x 40;
Both are factors of since the remainders are zero.
x
The remaining factors are x 1 and x 2.
2 f x
3x 2 x 1x 2
0
8
8
0
x 5, x 4
60. f x 8x Factors: x 2, x 4
4 14x3 71x2 10x 24;
(a)
2 8
4 8
8
8
14
16
30
30
32
2
71
60
11
11
8
3
10
22
12
12
(b) 8x
The remaining factors are 4x 3 and 2x 1.
2 2x 3 4x 32x 1
0
12
61. Factors:
(a) 1
f x 6x
2 6 41 9 14
3 19 14
6 38 28 0
3 41x2 9x 14;
2
3
6
6
38
4
42
28
28
0
24
24
Both are factors since the remainders are zero.
(c) f x x 72x 13x 2
0
2x 1, 3x 2
Section 2.3 Polynomial and Synthetic Division 175
(c)
(d) Zeros:
(e)
(c)
−4 3
(c)
(d) Zeros:
(e)
f x 3x 3 2x 2 19x 6
(d) Zeros:
3x 1x 3x 2
1
3, 3, 2
35
−10
f x x 1x 2x 5x 4
1, 2, 5, 4
20
−6 6
−180
f x 4x 32x 1x 2x 4
3
4, 1
2, 2, 4
(e) 40
−3 5
−380
(b)
f x
This shows that
x
f x
so
x 7.
2x 13x 2
The remaining factor is x 7.
1
2x 2
6x 42 6x 7
6x 7,
3
(d) Zeros:
(e)
7, 1 2
,
2 3
320
−9 3
−40

176 Chapter 2 Polynomial and Rational Functions
62. f x 10x
Factors: 2x 5, 5x 3
3 11x2 72x 45;
(a)
(c)
(d) Zeros:
(e)
5
2
3
5
10
10
(b)
fx
This shows that
x
f x
so
x 3.
2x 55x 3
The remaining factor is x 3.
5
2x 3
10 36 18
6 18
10 30 0
10x 30 10x 3
10x 3,
5
f x x 32x 55x 3
3, 5 3
,
2 5
100
−4 4
1
−80
43 1
1
11
25
36
3
3
0
0
43
43
72
90
48
0
48
18
48
48
45
45
64. f x x Factors: x 43 , x 3
(a) 3 1
3 3x2 48x 144;
(b) The remaining factor is x 43 .
0
0
144
144
0
63.
Factors:
(a)
Both are factors since the remainders are zero.
(b)
f x
This shows that
x
f x
so
x 5.
2x 1x 5
The remaining factor is x 5.
(c) f x x 5x 52x 1
1
f x 2x
2x 1, x 5
1
2 2 1 10 5
1 0 5
2 0 10 0
5 2 0 10
25 10
2 25 0
2x 25 2x 5
2x 5,
2x 5 3 x2 10x 5;
(d) Zeros:
(e)
(c)
5, 5, 1
2
14
−6 6
(d) Zeros:
(e)
−6
f x x 43 x 43 x 3
±43, 3
60
−8 8
65.
(a) The zeros of f are 2 and
(b) An exact zero is
(c)
f x x 2x
x 2x 5x 5
2 f x x
±2.236.
x 2.
2 1 2 5 10
2 0 10
1 0 5 0
5
3 2x2 5x 10 66.
(a) The zeros of g are
(b) is an exact zero.
(c)
fx x 4x
x 4x 2x 2
2 gx x
x 4, x 1.414,
x 4
4 1 4 2 8
4 0 8
1 0 2 0
2
3 4x 2 2x 8
−240
x 1.414.

67.
69.
71.
ht t 3 2t 2 7t 2
(a) The zeros of h are t 2, t 3.732, t 0.268.
(b) An exact zero is t 2.
(c) 2 1 2 7 2
2 8 2
1 4 1 0
x 4 6x 3 11x 2 6x
x 2 3x 2
1 1
1
2 1
ht t 2t 2 4t 1
By the Quadratic Formula, the zeros of t
are 2 ± 3. Thus,
ht t 2t 2 3t 2 3
2 4t 1
1
t 2t 2 3 t 2 3 .
4x 3 8x 2 x 3
2x 3
3
2
4
4
8
6
2
4x 3 8x 2 x 3
x 3
2
Thus, 4x3 8x 2 x 3
2x 3
6
1
5
5
2
3
11
5
6
6
6
0
x 4 6x 3 11x 2 6x
x 1x 2
73. (a) and (b)
1800
3
1200
—CONTINUED—
1
3
2
4x 2 2x 2 22x 2 x 1
13
3
3
0
x4 6x 3 11x 2 6x
x 1x 2
6
6
0
0
0
0
2x 2 x 1, x 3
2 .
0
0
0
x 2 3x, x 2, 1
Section 2.3 Polynomial and Synthetic Division 177
68.
70.
72.
f s s 3 12s 2 40s 24
(a) The zeros of f are s 6, s 0.764, s 5.236
(b) s 6 is an exact zero.
(c) 6 1 12 40 24
6 36 24
1 6 4 0
f s s 6s 2 6s 4
8 1
s 6s 3 5s 3 5
x 3 x 2 64x 64
x 8
1
1
8
7
x 3 x 2 64x 64
x 8
2 1
1
2 1
1
9
2
11
11
2
9
64
56
8
5
22
17
17
18
x 4 9x 3 5x 2 36x 4
x 2 4
64
64
0
x 4 9x 3 5x 2 36x 4
x 2 4
x , x 8
2 7x 8
1
36
34
2
2
2
0
x 4 9x 3 5x 2 36x 4
x 2x 2
4
4
0
x , x ±2
2 9x 1

178 Chapter 2 Polynomial and Rational Functions
73. —CONTINUED—
76. True.
78.
(c)
74. (a) and (b)
1
2
40
2
0
(b)
(c)
M 0.242t 3 12.43t 2 173.4t 2118 (d)
The model is a good fit to the actual data.
6
6
Year, t Military Personnel M
3 1705 1703
4 1611 1608
5 1518 1532
6 1472 1473
7 1439 1430
8 1407 1402
9 1386 1388
10 1384 1385
11 1385 1393
12 1412 1409
13 1434 1433
18 0.0026
For the year 2008, the model predicts a monthly rate
of about $49.38.
1
3
4
0.0026
92
2
90
12
R 0.0026t 3 0.0292t 2 1.558t 15.632
45
45
0
184
0
184
4
92
96
48
48
0
fx 2x 1x 1x 2x 33x 2x 4
fx x kqx r
0.0292
0.0468
0.0176
1.558
0.3168
1.8748
15.632
33.7464
49.3784
(a) any quadratic
where One example:
fx x 2x2 5 x3 2x2 ax
a > 0.
5
2 k 2, r 5, qx
bx c
18 0.242
0.242
12.43
4.356
8.074
M18 1613 thousand
173.4
145.332
28.068
2118
505.224
1612.776
No, this model should not be used to predict the
number of military personnel in the future. It
predicts an increase in military personnel until 2024
and then it decreases and will approach negative
infinity quickly.
75. False. If 7x 4 is a factor of f, then is a zero of f.
4
7
77. True. The degree of the numerator is greater than the
degree of the denominator.
(b) any quadratic
where One example:
fx x 3x2 1 x3 3x2 ax
a < 0.
1
2 k 3, r 1, qx
bx c

79.
83.
x n 3 ) x 3n 9x 2n 27x n 27
x 3n 9x 2n 27x n 27
x n 3
5 1
1
x 3n 3x 2n
4
5
9
6x 2n 27x n
6x 2n 18x n
3
45
42
x 2n 6x n 9
9x n 27
9x n 27
c
210
c 210
x 2n 6x n 9
81. A divisor divides evenly into a dividend if the remainder
is zero.
85.
87.
89.
To divide evenly, c 210 must equal zero. Thus, c must
equal 210.
fx x 3 2 x 3x 1 3
The remainder when k 3 is zero since x 3
is a factor of fx.
9x 2 25 0
3x 53x 5 0
5x 2 3x 14 0
5x 7x 2 0
91. 2 x2 6x 3 0
0
3x 5 0 ⇒ x 5
3
3x 5 0 ⇒ x 5
3
5x 7 0 ⇒ x 7
5
x 2 0 ⇒ x 2
x b ± b2 4ac
2a
3 ± 3
2
6 ± 62 423
22
Section 2.3 Polynomial and Synthetic Division 179
80.
x n 2 ) x 3n 3x 2n 5x n 6
x 3n 2x 2n
x 2n 5x n
x 2n 2x n
x 3n 3x 2n 5x n 6
x n 2
x 2n x n 3
3 x n 6
3 x n 6
0
x 2n x n 3
82. You can check polynomial division by multiplying the
quotient by the divisor. This should yield the original
dividend if the multiplication was performed correctly.
84.
2 1
0
2
0
4
2
8
1
20
c
42
1 2 4 10 21 c 42
To divide evenly, c 42 must equal zero. Thus, c must
equal 42.
86. In this case it is easier to evaluate f2 directly because
f x is in factored form. To evaluate using synthetic
division you would have to expand each factor and then
multiply it all out.
88. 16x2 21 0
90.
16x 2 21
4x 5 0
x 2 21
16
x 5
4
x ± 21
16
x ± 21
4
8x 2 22x 15 0
4x 52x 3 0
6 ± 12
4
or
or x 3
2x 3 0
2

180 Chapter 2 Polynomial and Rational Functions
92.
93.
95.
x 2 3x 3 0
x 3 ± 32 413
21
f x x 0x 3x 4
xx 3x 4 xx 2 7x 12
x 3 7x 2 12x
Note: Any nonzero scalar multiple of f x would also
have these zeros.
x 3x 12 2 2 f x x 3x 1 2 x 1 2
x 3x 1 2x 1 2
x 3x 2 2x 1
x 3 x 2 7x 3
Note: Any nonzero scalar multiple of f x would also
have these zeros.
Section 2.4 Complex Numbers
■ Standard form: a bi.
If b 0, then a bi is a real number.
If a 0 and b 0, then a bi is a pure imaginary number.
■ Equality of Complex Numbers: a bi c di if and only if a c and b d
■ Operations on complex numbers
(a) Addition:
(b) Subtraction:
(c) Multiplication:
a bi a bi c di ac bd
(d) Division:
c di c di c di c
■ The complex conjugate of a bi is a bi:
2 bc ad
d 2 c2 d 2i
a bia bi a 2 b 2
■ The additive inverse of a bi is a bi.
■ a ai for a > 0.
Vocabulary Check
3 ± 21
2
a bi c di a c b di
a bi c di a c b di
a bic di ac bd ad bci
1. (a) iii (b) i (c) ii 2.
94. f x x 6x 1
1; 1
x 6x 1
x 2 5x 6
3. principal square 4. complex conjugates
96.
Note: Any nonzero scalar multiple of f x would also
have these zeros.
x2 x 2x 22 3 2 f x x 1x 2x 2 3x 2 3
x 1x 2x 2 3x 2 3
x 2 x 2x 2 4x 1
x 4 3x 3 5x 2 9x 2
Note: Any nonzero scalar multiple of f x would also have
these zeros.

1.
4.
7.
a 10
b 6
Section 2.4 Complex Numbers 181
a bi 10 6i 2. a bi 13 4i 3. a 1 b 3i 5 8i
2b 5
b 5
2
a 6 6
a 0
a 13
b 4
a 1 5 ⇒ a 6
b 3 8 ⇒ b 5
a 6 2bi 6 5i 5. 4 9 4 3i 6. 3 16 3 4i
2 27 2 27i 8. 1 8 1 22i 9. 75 75i 53i
2 33i
10. 4 2i 11. 8 8 0i 8 12. 45
13. 6i i2 6i 1 14. 4i 2 2i 41 2i 15. 0.09 0.09 i
1 6i
4 2i
0.3i
16. 0.0004 0.02i 17. 5 i 6 2i 11 i 18. 13 2i 5 6i 8 4i
19.
21.
22.
8 i 4 i 8 i 4 i 20. 3 2i 6 13i 3 2i 6 13i
4
2 8 5 50 2 22i 5 52i
4
3 32i
8 18 4 32i 8 32i 4 32i 23.
24. 22 5 8i 10i 17 18i 25.
26. 1.6 3.2i 5.8 4.3i 4.2 7.5i 27.
28.
30.
6 2i2 3i 12 18i 4i 6i 2 29.
12 22i 6 6 22i
32 72i
3 11i
13i 14 7i 13i 14 7i
3
2
5
2i 53
14 20i
11
3 i 32
9
6
1 7
6 6i 1 i3 2i 3 2i 3i 2i 2
5 5 11
2i 3 3 i
3 i 2 5 i
6i5 2i 30i 12i 2 30i 12
12 30i
15 10 22
6 i 6 6 i
8i9 4i 72i 32i 2 31. 14 10 i14 10 i 14 10i2 14 10 24

182 Chapter 2 Polynomial and Rational Functions
32.
34.
3 15i3 15i 3 15i 2 33.
4 9 12i
5 12i
3 151
3 15 18
2 3i 2 4 12i 9i 2 35.
4 5i 2 16 40i 25i 2
16 40i 25
9 40i
2 3i 2 2 3i 2 4 12i 9i 2 4 12i 9i 2
4 12i 9 4 12i 9
10
36. 1 2i2 1 2i2 1 4i 4i 2 1 4i 4i 2 37. The complex conjugate of 6 3i is 6 3i.
1 4i 4i 2 1 4i 4i 2
8i
6 3i6 3i 36 3i 2 36 9 45
38. The complex conjugate of 7 12i is 7 12i.
39. The complex conjugate of 1 5i is 1 5i.
7 12i7 12i 49 144i 2
49 144
193
1 5i1 5i 1 2 5i 2
1 5 6
40. The complex conjugate of 3 2i is 3 2i. 41. The complex conjugate of 20 25i is 25i.
3 2i3 2i 9 2i 2
9 2
11
25i25i 20i 2 20
42. The complex conjugate of 15 15i is 15i. 43. The complex conjugate of 8 is 8.
15i15i 15i 2 15 15
44. The complex conjugate of 1 8 is 1 8.
45.
1 81 8 1 28 8
5
i
46. 14 2i
2i 2i
48.
50.
5 1 i
1 i 1 i
9 42
28i 28i
7i 47.
2 4i 4
5 5i 5 5i 5 5
2 1 i 2 2 2 i
6 7i
2
1 2i 6 12i 7i 14i
1 2i 1 2i 1 4i 2
20 5i
5
20
5
5
i 4 i
5
49.
51.
88 8
5 i 5i
i i 1 5i
2 2 4 5i
4 5i 4 5i 4 5i
24 5i
16 25
3 i 3 i 3 i
3 i 3 i 3 i
9 6i i 2
9 1
6 5i 6 5i
i i
i
i
8 10i
41
8 10
41 41 i
8 6i 4 3
10 5 5 i
6i 5i 2
5 6i
1

52.
53.
55.
57.
59.
8 16i
2i
3i
4 5i2
2i 16i 32i2
2i 4i2 8 4i
3i
16 40i 25i2
27i 120i2
81 1600
120 27
1681 1681 i
3i 9 40i
9 40i 9 40i
120 27i
1681
2 3 21 i 31 i
1 i 1 i 1 i1 i
2 2i 3 3i
1 1
1 5
2 2 i
1 5i
2
i 2i i3 8i 2i3 2i
3 2i 3 8i 3 2i3 8i
3i 8i2 6i 4i 2
9 24i 6i 16i 2
4i2 9i
9 18i 16
4 9i 25 18i
25 18i 25 18i
100 72i 225i 162i2
625 324
100 297i 162
949
62 297i
949
62 297
949 949 i
6 2 6i2i 12i 2 23 1 60.
23
54.
56.
58.
Section 2.4 Complex Numbers 183
5i
5i
2
2 3i 4 12i 9i2 5i 5 12i
5 12i 5 12i
25i 60i2
25 144i 2
60 25i
169
12
5
12 9i
5
60 25
169 169 i
2i 5 2i2 i
2 i 2 i 2 i2 i
52 i
2 i2 i
4i 2i 2 10 5i
4 i 2
9
5 i
1 i 3 1 i4 i 3i
i 4 i i4 i
4 i 4i i 2 3i
4i i 2
5 1 4i
1 4i 1 4i
5 20i
1 16i 2
5 20
17 17 i
5 10 5i10i
50i 2 521 52
61. 10 2 10i 2 10i 2 10 62. 75 2 75i 2 75i 2 75
63. 3 5 7 10 3 5i7 10i
21 310i 75i 50i 2
21 50 75 310 i
21 52 75 310 i

184 Chapter 2 Polynomial and Rational Functions
64.
66.
2 6 2 2 6i2 6i 65.
6 ± 4
2
3 ± i
4 26i 26i 6i 2
4 26i 26i 61
4 6 46i
2 46i
x a 1, b 6, c 10
2 6x 10 0; 67.
x 6 ± 62 4110
21
x 2 2x 2 0; a 1, b 2, c 2
x 2 ± 22 412
21
2 ± 4
2
2 ± 2i
2
1 ± i
4x 2 16x 17 0; a 4, b 16, c 17
x 16 ± 162 4417
24
16 ± 16
8
16 ± 4i
8
2 ± 1
2 i
68. 9x2 6x 37 0; a 9, b 6, c 37 69. 4x2 16x 15 0; a 4, b 16, c 15
x 6 ± 62 4937
29
6 ± 1296
18
1 36i 1
± ± 2i
3 18 3
4 ± 176
32
4 ± 411i
32
1 11
±
8 8 i
x 16 ± 162 4415
24
16 ± 16
8
x 12
8 3
2
12 ± 72
6
12 ± 62i
6
16 ± 4
8
or x 20
8 5
2
70.
t
4 ± 42 16t
4163
216
2 4t 3 0; a 16, b 4, c 3 71. Multiply both sides by 2.
x 12 ± 122 3x
4318
23
2 12x 18 0
3
2 x2 6x 9 0
12 ± 136
28
12 ± 2i34
28
3 34
±
7 14 i
10 ± 1500
14
5 ± 515
7
2 ± 2i
72.
x 12 ± 122 14x
4145
214
2 12x 5 0; a 14, b 12, c 5
7
8 x2 3 5
x 0
4 16
73. Multiply both sides by 5.
x 10 ± 102 7x
4750
27
2 1.4x
10x 50 0
2 2x 10 0
10 ± 1015
14
5 515
±
7 7

74. 4.5x2 3x 12 0; a 4.5, b 3, c 12 75. 6i 3 i 2 6i2i i2 x 3 ± 32 44.512
24.5
3 ± 207
9
3 ± 3i23
9
1 23
±
3 3 i
76. 4i 2 2i 3 4 2i 77.
79.
75 3 53i 3
5 3 3 3 i 3
12533 1i
3753i
81. 1 1 1 i
i3
i i i
83. (a)
(b)
84. (a)
(b)
(c)
z 1 9 16i, z 2 20 10i
1 1
z z1 1
z2 2 3 8
511i 5i
i i
i 82.
i2 1
1
9 16i
1
20 10i
z 340 230i 29 6i
29 6i 29 6i
1 3i 3 1 3 31 2 3i 313i 2 3i 3
1 33i 9i 2 33i 3
1 33i 9 33i
8
1 33i 9i 2 33i 3
1 33 i 9 33i
8
Section 2.4 Complex Numbers 185
61i 1
6i 1
1 6i
5i 5 5i 2 i 2 i 78. i 3 1i 3 1i i
80. 2 6 2i 6 8i 6 8i 4 i 2 8
1
2i
1 1
3 8i 3 8i
1 3i 3 1 3 31 2 3i 313i 2 3i 3
85. (a)
(b)
(c) 2i4 24i 4 2
161 16
4 2
16
4 16
(d) 2i 4 2 4 i 4 161 16
20 10i 9 16i
9 16i20 10i
11,240 4630i
877
11,240
877
4630
877 i
86. (a)
87. False, if b 0 then a bi a bi a.
88. True
That is, if the complex number is real, the number equals
its conjugate.
29 6i
340 230i
(b)
i 40 i 4 10 1 10 1
i 25 i 4 6 i 1 6 i i
8i 8i 1
8i 64i 2 8 i
(c) i50 i 412i 2 11 1
(d) i 67 i 4 16 i 3 1i i
i6 4
i6 2
x 4 x 2 14 56
14 ? 56
36 6 14 ? 56
56 56

186 Chapter 2 Polynomial and Rational Functions
89. False
i 44 i 150 i 74 i 109 i 61 i 4 11 i 4 37 i 2 i 4 18 i 2 i 4 27 i i 4 15 i
90. 66 6i6i 6i 2 6
91.
92.
1 11 1 37 1 1 18 1 1 27 i 1 15 i
1 1 1 i i 1
a 1 b 1ia 2 b 2i a 1a 2 a 1b 2i a 2b 1i b 1b 2i 2
a 1a 2 b 1b 2 a 1b 2 a 2b 1i
The complex conjugate of this product is a1a2 b1b2 a1b2 a2b1i. The product of the complex conjugates is:
a 1 b 1 ia 2 b 2 i a 1 a 2 a 1 b 2 i a 2 b 1 i b 1 b 2 i 2
a 1 a 2 b 1 b 2 a 1 b 2 a 2 b 1i
Thus, the complex conjugate of the product of two complex numbers is the product of their complex conjugates.
a 1 b 1i a 2 b 2i a 1 a 2 b 1 b 2i
The complex conjugate of this sum is a1 a2 b1 b2i. The sum of the complex conjugates is a1 b1i a2 b2i a1 a2 b1 b2i. Thus, the complex conjugate of the sum of two complex numbers is the sum of their complex conjugates.
93. 4 3x 8 6x x 2 x 2 3x 12
94. x3 3x2 6 2x 4x2 x3 3x2 6 2x 4x2 95. 3x 1
2x 4 3x2 12x 1
2x 2
4x 2 20x 25
x 3 x 2 2x 6
x 31
x 31
3x2 23
2 x 2
96. 2x 52 2x2 22x5 52 97. x 12 19 98. 8 3x 34
99. 45x 6 36x 1 0 100. 5x 3x 11 20x 15 101.
102.
20x 24 18x 3 0
r 2 m 1 m 2
F
2x 27 0
2x 27
x 27
2
r m1m2 F m1m2
F
F
F m1m2F F
5x 15x 55 20x 15
30x 40
x 40
30 4
3
3x 42
x 14
V 4
3 a2 b
3V 4a 2 b
3V
a2
4b
3V
a
4b
F m1m2 r2 103. Let x #
liters withdrawn and replaced.
0.505 x 1.00x 0.605
2.50 0.50x 1.00x 3.00
0.50x 0.50
x 1 liter
a 1
2 3V 3Vb
b 2b

Section 2.5 Zeros of Polynomial Functions
Section 2.5 Zeros of Polynomial Functions 187
■ You should know that if f is a polynomial of degree then f has at least one zero in the complex number system.
■ You should know the Linear Factorization Theorem.
■ You should know the Rational Zero Test.
■ You should know shortcuts for the Rational Zero Test. Possible rational zeros
(a) Use a graphing or programmable calculator.
(b) Sketch a graph.
(c) After finding a root, use synthetic division to reduce the degree of the polynomial.
■ You should know that if is a complex zero of a polynomial f, with real coefficients, then is also a complex
zero of f.
■ You should know the difference between a factor that is irreducible over the rationals (such as ) and a factor that is
irreducible over the reals (such as x ).
■ You should know Descartes’s Rule of Signs. (For a polynomial with real coefficients and a non-zero constant term.)
(a) The number of positive real zeros of f is either equal to the number of variations of sign of f or is less than
that number by an even integer.
(b) The number of negative real zeros of f is either equal to the number of variations in sign of fx or is less than
that number by an even integer.
(c) When there is only one variation in sign, there is exactly one positive (or negative) real zero.
■ You should be able to observe the last row obtained from synthetic division in order to determine upper or lower bounds.
(a) If the test value is positive and all of the entries in the last row are positive or zero, then the test value is an upper
bound.
(b) If the test value is negative and the entries in the last row alternate from positive to negative, then the test value is a
lower bound. (Zero entries count as positive or negative.)
2 x
9
2 n > 0,
factors of constant term
factors of leading coefficient
a bi
a bi
7
Vocabulary Check
1.
3.
1. Fundamental Theorem of Algebra 2. Linear Factorization Theorem 3. Rational Zero
4. conjugate 5. irreducible; reals 6. Descarte’s Rule of Signs
7. lower; upper
f x xx 6 2 2.
The zeros are: x 0, x 6
gx x 2x 4 3 4.
The zeros are: x 2, x 4
6. ht t 3t 2t 3it 3i
The four zeros are: 3, 2, ±3i
7.
f x x 2 x 3x 2 1 x 2 x 3x 1x 1
The five zeros are: 0, 0, 3, ±1
f x x 5x 82 5. f x x 6x ix i
The three zeros are: 5, 8, 8
fx x 3 3x 2 x 3
Possible rational zeros: ±1, ±3
The three zeros are:
x 6, x i, x i
Zeros shown on graph: 3, 1, 1

188 Chapter 2 Polynomial and Rational Functions
8.
10.
12.
16.
f x x 3 4x 2 4x 16
Possible rational zeros: ±1, ±2, ±4, ±8, ±16
Zeros shown on graph: 2, 2, 4
Possible rational zeros:
Zeros shown on graph: 1, 1
±1, ±2, ±
1
2 , 2 , 1, 2
1
f x 4x
1
2 , ± 4
5 8x 4 5x 3 10x 2 x 2
f x x 3 7x 6 13.
Possible rational zeros: ±1, ±2, ±3, ±6
3 1
1
14. hx x 3 9x 2 20x 12
Possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12
1 1
1
hx x 1x 2 8x 12
Thus, the rational zeros are 1, 2, 6.
px x 3 9x 2 27x 27
Possible rational zeros: ±1, ±3, ±9, ±27
3 1
x 1x 2x 6
1
0
3
3
9
1
8
9
3
6
7
9
2
20
8
12
27
18
f x x 3x 2 6x 9
x 3x 3x 3
Thus, the rational zero is 3.
9
6
6
0
f x x 3x 2 3x 2 x 3x 2x 1
Thus, the rational zeros are 2, 1, 3.
12
12
0
27
27
0
9.
11.
15. ht t3 12t2 21t 10
17.
fx 2x 4 17x 3 35x 2 9x 45
Possible rational zeros:
Zeros shown on graph: 1, 3
2 , 3, 5
fx x 3 6x 2 11x 6
Possible rational zeros: ±1, ±2, ±3, ±6
1 1
1
x 3 6x 2 11x 6 x 1x 2 5x 6
Thus, the rational zeros are 1, 2, and 3.
Possible rational zeros: ±1, ±2, ±5, ±10
1 1
1
t 3 12t 2 21t 10 t 1t 2 11t 10
Thus, the rational zeros are 1 and 10.
Cx 2x 3 3x 2 1
Possible rational zeros:
1 2
2
6
1
5
12
1
11
3
2
1
11
5
6
21
11
10
0
1
1
±1, ±3, ±5, ±9, ±15, ±45,
± 1 3 5 9 15 45
2 , ± 2 , ± 2 , ± 2 , ± 2 , ± 2
6
6
0
x 1x 2x 3
gx x 3 4x 2 x 4 x 2 x 4 1x 4
x 4x 2 1
x 4x 1x 1
Thus, the rational zeros of gx are 4 and ±1.
10
10
0
t 1t 1t 10
t 1 2 t 10
±1, ± 1
2
1
1
0
2x 3 3x 2 1 x 12x 2 x 1
x 1x 12x 1
x 1 2 2x 1
Thus, the rational zeros are and 1
2 . 1

18.
19.
f x 3x 3 19x 2 33x 9
Possible rational zeros:
3 3
3
f x x 33x 2 10x 3 x 33x 1x 3
Thus, the rational zeros are 3, 1
3 .
fx 9x 4 9x 3 58x 2 4x 24
Possible rational zeros:
2 9
9
3 9
9
19
9
10
9
18
27
27
27
0
33
30
58
54
4
4
0
4
4
8
12
12
12
0
9x 4 9x 3 58x 2 4x 24
3
±1, ±3, ±9, ± 1
3
9
9
0
±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24,
± 1 2 4 8 1 2 4 8
3 , ± 3 , ± 3 , ± 3 , ± 9 , ± 9 , ± 9 , ± 9
24
24
0
x 2x 39x 2 4
x 2x 33x 23x 2
Thus, the rational zeros are and ± 2
3 .
2, 3,
21. z4 z3 2z 4 0
23.
Possible rational zeros: ±1, ±2, ±4
1 1
1
2 1
1
Possible rational zeros:
1 2
2
6 2
2
1
1
2
2
2
0
7
2
9
9
12
3
0
2
2
2
0
2
26
9
17
17
18
1
2
2
4
4
4
0
23
17
6
6
6
0
4
4
0
z 4 z 3 2z 4 z 1z 2z 2 2
The only real zeros are 1 and 2.
2y 4 7y 3 26y 2 23y 6 0
The only real zeros are 1, 6, and 1
2.
±1, ±2, ±3, ±6, ± 1 3
2 , ± 2
6
6
0
20.
22.
Section 2.5 Zeros of Polynomial Functions 189
f x 2x 4 15x 3 23x 2 15x 25
Possible rational zeros:
5 2
1 2
2
2
1 2
2
15
10
5
5
2
3
3
2
5
23
25
2
2
3
5
5
5
0
±1, ±5, ±25, ± 1 5 25
2 , ± 2 , ± 2
15
10
5
5
5
0
25
25
0
f x x 5x 1x 12x 5
Thus, the rational zeros are 5, 1, 1, 5
2 .
x 4 13x 2 12x 0
xx 3 13x 12 0
Possible rational zeros of x
±1, ±2, ±3, ±4, ±6, ±12
3 13x 12:
1 1
1
0
1
1
13
1
12
12
12
0
xx 1x 2 x 12 0
xx 1x 4x 3 0
The real zeros are 0, 1, 4, 3.
y 1y 62y 1y 1 y 12 2y y 62y 1
4 7y3 26y2 23y 6 y 1y 62y2 3y 1

190 Chapter 2 Polynomial and Rational Functions
24.
26.
28.
30.
x 5 x 4 3x 3 5x 2 2x 0
xx 4 x 3 3x 2 5x 2 0
Possible rational zeros of x
±1, ±2
4 x3 3x2 5x 2:
1 1
f x 4x 3 12x 2 x 15
(a) Possible rational zeros:
(b)
15
12
−9 −6 −3 6 9 12
(c) Real zeros: 1, 3 5
2 , 2
f x 4x 4 17x 2 4
(a) Possible rational zeros:
(b)
1
2 1
1
1
1
0
0
2
2
xx 1x 2x 2 2x 1 0
xx 1x 2x 1x 1 0
The real zeros are 2, 0, 1.
(c) Real zeros: ±2, ± 1
2
y
3
0
3
−8 8
9
−15
3
4
1
5
3
2
2
2
0
2
2
0
± 5
±1, ±3, ±5, ±15, ±
15 1 3 5 15
2 , ± 2 , ± 4 , ± 4 , ± 4 , ± 4
1 3
2 , ± 2 ,
x
±1, ±2, ±4, ± 1 1
2 , ± 4
25.
29.
fx x 3 x 2 4x 4
(a) Possible rational zeros:
(b)
(c) The zeros are: 2, 1, 2
fx 2x 4 13x 3 21x 2 2x 8
(a) Possible rational zeros:
(b)
16
−4 8
−8
(c) The zeros are: 1
2 , 1, 2, 4
31. fx 32x
(a) Possible rational zeros:
3 52x2 17x 3
(b)
−6 −4 4 6
f x 3x 3 20x 2 36x 16 27. fx 4x3 15x2 8x 3
(a) Possible rational zeros:
(b)
10
8
6
4
2
−4 −2 6 8 10 12
−4
−6
y
(c) Real zeros: 2
3 , 2, 4
± 2
±1, ±2, ±4, ±8, ±16, ±
4 8 16
3 , ± 3 , ± 3 , ± 3
1
3 ,
x
6
−1 3
−2
4
2
−4
−6
−8
(a) Possible rational zeros:
(b)
4
2
(c) The zeros are: 1
8, 3
4, 1
y
−6 −4 −2 2 4 6 8 10
−4
−6
y
(c) The zeros are: 1
4 , 1, 3
±1, ±2, ±4
x
±1, ±3, ± 1
2, ± 3
2, ± 1
4, ± 3
4
x
±1, ±2, ±4, ±8, ± 1
2
± 1
±1, ±3, ±
3 1 3 1 3
6 , ± 8 , ± 16 , ± 16 , ± 32 , ± 32
1 3 1 3
2 , ± 2 , ± 4 , ± 4 ,

−24
(c) Real zeros: 2, 1 145
±
8 8
Section 2.5 Zeros of Polynomial Functions 191
32.
(a) Possible rational zeros:
±
(b)
8
−8 8
1
f x 4x
±1, ±2, ±3, ±6, ±9, ±18,
3 9 1 3 9
2 , ± 2 , ± 2 , ± 4 , ± 4 , ± 4
3 7x 2 11x 18 33. fx x
(a) From the calculator we have x ±1 and x ±1.414.
(b) An exact zero is x 1.
1 1 0 3 0 2
1 1 2 2
1 1 2 2 0
4 3x2 2
34. Pt t 4 7t 2 12
(a)
(b) 2 1
(c)
(c)
t ±2, ±1.732
1
2 1
1
0
2
2
7
4
3
2
2
0
0
6
6
3
0
3
6
6
0
Pt t 2t 2t 2 3
12
12
0
t 2t 2t 3 t 3
3 6
6
7
18
11
30
33
3
9
9
0
gx x 3x 36x 2 11x 3
x 3x 33x 12x 3
(c) 1 1
1
1
1
0
2
0
2
2
2
0
fx x 1x 1x 2 2
x 1x 1x 2 x 2
35.
(a) hx xx4 7x3 10x2 hx x
14x 24
5 7x4 10x3 14x2 24x
From the calculator we have x 0, 3, 4 and x ±1.414.
(b) An exact zero is x 3.
(c)
3 1
1
4 1
1
7
3
4
4
4
0
10
12
2
2
0
2
14
6
8
8
8
0
hx xx 3x 4x 2 2
24
24
0
xx 3x 4x 2 x 2
36. gx 6x
(a) x ±3, 1.5, 0.333
(b) 3 6 11 51 99 27
18 21 90 27
6 7 30 9 0
4 11x 3 51x 2 99x 27 37.
Note: f x ax where a is any
nonzero real number, has the zeros 1 and ±5i.
3 x2 x
25x 25,
3 x2 x 1x
25x 25
2 f x x 1x 5ix 5i
25
38.
f x x 4x 3ix 3i 39. f x x 6x 5 2ix 5 2i
x 4x 2 9
x 3 4x 2 9x 36
Note: f x ax where a is
any real number, has the zeros 4, 3i and 3i.
3 4x 2 9x 36,
x 6x 5 2ix 5 2i
x 6x 5 2 2i 2
x 6x 2 10x 25 4
x 6x 2 10x 29
x 3 4x 2 31x 174
Note: f x ax where a is any
nonzero real number, has the zeros 6, and 5 ± 2i.
3 4x2 31x 174,

192 Chapter 2 Polynomial and Rational Functions
40.
45.
f x x 2x 4 ix 4 i
x 2x 2 8x 17
x 3 10x 2 33x 34
Note: f x ax where a is
any real number, has the zeros 2, 4 ± i.
3 10x 2 33x 34
42. If 1 3i is a zero, so is its conjugate, 1 3i.
44.
f x x 5 2 x 1 3ix 1 3i
x 2 10x 25x 2 2x 4
x 4 8x 3 9x 2 10x 100
Note: f x ax where
a is any real number, has the zeros 5, 5, 1 ± 3i.
4 8x 3 9x 2 10x 100,
f x x 4 2x 3 3x 2 12x 18
x 4
2x 3
6x 2
x 2 2x 3
x 6 ) x 4 2x 3 3x 2 12x 18
x 4 2x 3 3x 2 12x 6
12x
3x2 3x
0
2 2x
18
3 3x2 18
f x x 4 4x 3 5x 2 2x 6
x 2 2x 2 ) x 4 4x 3 5x 2 2x 6
x 4 2x 3 2x 2
x 4 2x 3 7x 2 2x 6
2x 3 4x 2 4x
2x 3 3x 2 6x 6
3x 2 6x 6
3x 2 6x 0
fx x 2 2x 2x 2 2x 3
46. fx x4 3x3 x2 12x 20
x 2 2x 3
x2 4 ) x4 3x3 x2 x
12x 20
2 3x 5
x 4 4x 2
3x 3 5x 2 12x
3x 3 12x
5x2 5x
20
2 20
0
41. If is a zero, so is its conjugate,
3x2 x 2x 32 2i 2 3 2i
3 2i.
f x 3x 2x 1x 3 2ix 3 2i
3x 2x 1x 3 2ix 3 2i
3x 2 x 2x 2 6x 9 2
3x 2 x 2x 2 6x 11
3x 4 17x 3 25x 2 23x 22
Note: f x a3x where a is
2
any nonzero real number, has the zeros , 1, and 3 ±2i.
4 17x3 25x2 23x 22,
43.
(a)
(b) f x x
(c) f x x 3ix 3ix 3x 3
2 f x x
9x 3x 3
2 9x2 f x x
3
4 6x2 27
(a)
(b) f x x 6x 6x 2 f x x
2x 3
2 6x 2 2x 3
(c) f x x 6x 6x 1 2ix 1 2i
(a)
(b) fx x 1 3 x 1 3 x
(c) fx x 1 3 x 1 3 x 1 2 i x 1 2 i
Note: Use the Quadratic Formula for (b) and (c).
2 fx x
2x 3
2 2x 2x2 2x 3
(a)
(b)
f x x 2 4x 2 3x 5
f x x 2 3 29
4x
2
3 29
(c) f x x 2ix 2ix
2
3 29
x
2
3
3 29
x
2

47.
48.
fx 2x 3 3x 2 50x 75
Since 5i is a zero, so is 5i.
5i 2
2
5i 2
The zero of The zeros of
are x 3
2x 3 is x fx
2 and x ±5i.
3
2 .
49. fx 2x
Since 2i is a zero, so is 2i.
4 x3 7x2 4x 4
2i 2
2
2
2i 2
2
3
10i
3 10i
3 10i
10i
3
1
4i
1 4i
1 4i
4i
1
50
50 15i
15i
15i
15i
0
f x x 3 x 2 9x 9
Since 3i is a zero, so is 3i.
3i 1
1
3i 1
1
1
3i
1 3i
1 3i
3i
1
9
9 3i
3i
3i
7
8 2i
1 2i
1 2i
2i
1
75
75
0
4
4 2i
2i
2i
2i
0
The zeros of
are and The zeros of are
x ±2i, x and x 1.
1
2 ,
x x 1.
fx
1
2x
2
2 x 1 2x 1x 1
0
3i
9
9
0
4
4
0
Alternate Solution
The zero of x 1 is x 1. The zeros of f are x 1 and x ±3i.
50. gx x 3 7x 2 x 87
Since 5 2i is a zero, so is 5 2i.
5 2i 1
1
5 2i 1
1
7
5 2i
2 2i
2 2i
5 2i
3
1
14 6i
15 6i
15 6i
15 6i
Section 2.5 Zeros of Polynomial Functions 193
Since are zeros of fx, x 5ix 5i x is a
factor of fx. By long division we have:
2 x ±5i
25
02x 3
x 2 0x 25 ) 2x 3 3x 2 50x 75
2x 3 0x 2 50x
2x 3 3x 2 50x 75
3x 2 50x 75
3x 2 50x 70
Thus, and the zeros of f are and x 3
x ±5i
fx x2 252x 3
Alternate Solution
Since are zeros of fx, x 2ix 2i x is a factor
of fx. By long division we have:
2 x ±2i
4
x 2 0x 4 ) 2x 4 x 3 7x 2 4x 4
Thus,
The zero of x 3 is x 3. The zeros of f are x 3, 5 ± 2i.
0
87
87
0
2x 4 0x 3 8x 2
2 x 2 x 1
x 3 x 2 4x
x 3 0x 2 4x
x 2 0x 4
x 2 0x 4
fx x 2 42x 2 x 1
fx x 2ix 2i2x 1x 1
and the zeros of are x ±2i, x 1
, and x 1.
fx
0
2
2 .

194 Chapter 2 Polynomial and Rational Functions
51. gx 4x
Since 3 i is a zero, so is 3 i.
3 23x2 34x 10
3 i 4
4
3 i 4
4
23
12 4i
11 4i
11 4i
12 4i
1
34
37 i
3 i
3 i
3 i
0
10
10
0
The zero of The zeros of are
x 3 ± i and x 1
4 .
4x 1 is x gx
1
4 .
52. hx 3x3 4x 2 8x 8
53.
Since 1 3 i is a zero, so is 1 3i.
1 3i 3
3
1 3i 3
3
4
3 33i
1 33i
1 33i
3 33i
2
The zero of The zeros of f are x 2
3x 2 is x 2
, 1 ± 3i.
fx x 4 3x 3 5x 2 21x 22
3 .
8
10 23i
2 23i
2 23i
2 23i
Since 3 2 i is a zero, so is 3 2 i, and
is a factor of fx. By long division, we have:
x 2 6x 11 ) x 4 3x 3 5x 2 21x 22
Thus,
x 4 6x 3 11x 2
3x 3 16x 2 21x
3x 3 18x 2 33x
f x x 2 6x 11x 2 3x 2
x 2 6x 11x 1x 2
0
x 2 3x 2
2x 2 12x 22
2x 2 12x 22
8
8
x 3 2 i x 3 2 i x 3 2 ix 3 2 i
and the zeros of f are x 3 ± 2 i, x 1, and x 2.
0
0
x 3 2 2 i 2
x 2 6x 11
3
Alternate Solution
Since 3 ± i
are zeros of gx,
x 32 i2 x 3 i x 3 i x 3 ix 3 i
x
is a factor of gx. By long division we have:
2 6x 10
34x 1
x 2 6x 10 ) 4x 3 23x 2 34x 10
4x 3 24x 2 40x
4x 3 24 x 2 36x 10
x 2 36x 10
Thus, and the zeros of
are x 3 ± i and x 1
gx x gx
4.
2 6x 104x 1
0

54. f x x
Since 1 3i is a zero, so is 1 3i.
3 4x 2 14x 20 55. fx x
x 5ix 5i
2 25
1 3i 1
1
1 3i 1
1
4
1 3i
3 3i
3 3i
1 3i
The zero of x 2 is x 2.
The zeros of f are x 2, 1 ± 3i.
62. hx x 3 3x 2 4x 2
Possible rational zeros: ±1, ±2
2
14
12 6i
2 6i
2 6i
2 6i
0
20
20
By the Quadratic Formula, the zeros of x
2 ± 4 8
are x 1 ± i.
2
Zeros: x 1, 1 ± i
hx x 1x 1 ix 1 i
2 1 1 3 4 2
1 2 2
1 2 2 0
2x 2
0
Section 2.5 Zeros of Polynomial Functions 195
The zeros of fx are x ±5i.
56. f x x
By the Quadratic Formula, the zeros of f x are
2 x 56 57. hx x
By the Quadratic Formula, the zeros of hx are
2 4x 1
1 ± 1 224 1 ± 223i
x .
2
2
1 223i 1 223i
f x x
2 x
2
58. gx x
By the Quadratic Formula, the zeros of f x are
2 10x 23 59.
x2 9x2 fx x
9
4 81
60.
x
10 ± 100 92
2
10 ± 8
2
gx x 5 2x 5 2
Zeros: y ±5, ±5i
f y y 5y 5y 5iy 5i
5 ± 2.
x
4 ± 16 4
2
2 ± 3.
hx x 2 3 x 2 3
x 2 3 x 2 3
x 3x 3x 3ix 3i
The zeros of fx are x ±3 and x ±3i.
y2 25y2 f y y
25
4 625 61. fz z
By the Quadratic Formula, the zeros of f z are
2 2z 2
z
2 ± 4 8
2
1 ± i.
fz z 1 iz 1 i
z 1 iz 1 i
63. gx x3 6x2 13x 10
Possible rational zeros:
2 1
By the Quadratic Formula, the zeros of x are
2 4x 5
x
1
6
2
4
13
8
5
4 ± 16 20
2
±1, ±2, ±5, ±10
10
10
0
2 ± i.
The zeros of gx are x 2 and x 2 ± i.
gx x 2x 2 ix 2 i
x 2x 2 ix 2 i

196 Chapter 2 Polynomial and Rational Functions
64.
65.
67.
f x x 3 2x 2 11x 52
Possible rational zeros: ±1, ±2, ±4, ±13, ±26
4 1
hx x 3 x 6
Possible rational zeros:
2 1
By the Quadratic Formula, the zeros of x are
2 2x 3
x
1
0
2
2
2 ± 4 12
2
1
4
3
±1, ±2, ±3, ±6
6
6
0
1 ± 2 i.
The zeros of hx are x 2 and x 1 ± 2 i.
hx x 2x 1 2 i x 1 2 i
x 2x 1 2 ix 1 2 i
fx 5x 3 9x 2 28x 6
Possible rational zeros:
±1, ±2, ±3, ±6, ± 1 2 3 6
5 , ± 5 , ± 5 , ± 5
1
5
By the Quadratic Formula, the zeros of
5x are
2 10x 30 5x2 2x 6
x
5
5
1
2
4
6
9
1
10
11
24
2 ± 4 24
2
13
28
2
30
52
52
By the Quadratic Formula, the zeros of x are x 2 6x 13
Zeros: x 4, 3 ± 2i
f x x 4x 3 2ix 3 2i
0
6
6
0
1 ± 5 i.
The zeros of are
fx x
5x 1x 1 5 ix 1 5 i
1
x
55x 1 5 i x 1 5 i
1
5 and x 1 ± 5 i.
fx
6 ± 36 52
2
66. hx x
Possible rational zeros: ±1, ±5, ±7, ±35
3 9x2 27x 35
By the Quadratic Formula, the zeros of x
4 ± 16 28
are x 2 ± 3i.
2
Zeros: 5, 2 ± 3i
hx x 5x 2 3ix 2 3i
2 5 1 9 27 35
5 20 35
1 4 7 0
4x 7
68. gx 3x 3 4x 2 8x 8
Possible rational zeros:
±1, ±2, ±4, ±8, ± 1 2 4 8
3 , ± 3 , ± 3 , ± 3
2
3
By the Quadratic Formula, the zeros of
3x are
2 6x 12 3x2 2x 4
x
3
3
3 ± 2i.
4
2
6
2 ± 4 16
2
8
4
12
8
8
0
1 ± 3i.
Zeros: x
gx 3x 2x 1 3ix 1 3i
2
3 , 1 ± 3i

69.
71.
73.
gx x 4 4x 3 8x 2 16x 16
Possible rational zeros: ±1, ±2, ±4, ±8, ±16
2 1
1
2 1
1
4
2
2
2
2
0
8
4
4
4
0
4
8
8
0
16
8
8
gx x 2x 2x 2 4 x 2 2 x 2ix 2i
The zeros of gx are 2 and ±2i.
fx x 4 10x 2 9
x 2 1x 2 9
16
16
0
x ix ix 3ix 3i
The zeros of fx are x ±i and x ±3i.
fx x 3 24x 2 214x 740 74.
Possible rational zeros:
Based on the graph, try x 10.
By the Quadratic Formula, the zeros of x are
2 14x 74
x
−20
10 1
1
2000
−1000
24
10
14
14 ± 196 296
2
±1, ±2, ±4, ±5, ±10, ±20, ±37,
±74, ±148, ±185, ±370, ±740
The zeros of fx are x 10 and x 7 ± 5i.
10
214
140
74
740
740
0
7 ± 5i.
Section 2.5 Zeros of Polynomial Functions 197
70. hx x
Possible rational zeros: ±1, ±3, ±9
4 6x 3 10x 2 6x 9
72.
3 1
1
3 1
1
6
3
3
3
3
0
10
9
1
0
The zeros of x are x ±i.
2 1
Zeros:
hx x 3 2 x 3, ±i
x ix i
1
Zeros: x ±2i, ±5i
1
6
3
3
3
f x x 4 29x 2 100
x 2 25x 2 4
0
3
9
9
f x x 2ix 2ix 5ix 5i
f s 2s 3 5s 2 12s 5
Possible rational zeros:
Based on the graph, try
1
2
By the Quadratic Formula, the zeros of 2s are
2 2s 5
s
2
2
10
−10 10
−10
5
1
4
12
2
10
2 ± 4 20
2
5
5
0
s 1
2 .
1 ± 2i.
The zeros of are s 1
and s 1 ± 2i.
fs
2
0
±1, ±5, ± 1 5
2 , ± 2

198 Chapter 2 Polynomial and Rational Functions
75.
78.
fx 16x 3 20x 2 4x 15 76.
Possible rational zeros:
± 5
4, ± 15
4 , ± 1
8, ± 3
8, ± 5
8, ± 15
8 , ± 1
16, ± 3
16, ± 5
16, ± 15
±1, ±3, ±5, ±15, ±
16
1
2, ± 3
2, ± 5
2, ± 15
2 , ± 1
4, ± 3
4,
Based on the graph, try
By the Quadratic Formula, the zeros of
16x are
2 32x 20 44x2 8x 5
x
−3
3
4
8 ± 64 80
8
The zeros of are x 3
fx
−4
16
16
gx x 5 8x 4 28x 3 56x 2 64x 32
Possible rational zeros: ±1, ±2, ±4, ±8, ±16, ±32
10
−10 10
−10
20
−5
20
12
32
−5
x 3
4.
4
24
20
1 ± 1
2 i.
Based on the graph, try x 2.
3
4
4
15
15
0
and x 1 ± 1
2 i.
2 1
2 1
2 1
By the Quadratic Formula, the zeros of x are
2 2x 4
x
1
1
1
f x 9x 3 15x 2 11x 5
Possible rational zeros:
−5 5
Based on the graph, try x 1.
1 9
By the Quadratic Formula, the zeros of 9x are
2 6x 5
x
The zeros of are x 1 and x 1
fx
8
2
6
6
2
4
4
2
2
9
28
12
16
8
8
4
4
2 ± 4 16
2
15
9
6
6 ± 36 180
18
16
8
5
−5
24
16
8
8
8
0
11
6
5
56
32
24
1 ± 3i.
±1, ±5, ± 1 5 1 5
3 , ± 3 , ± 9 , ± 9
5
5
0
1 2
±
3 3 i.
77.
Possible rational zeros: ±1, ±2, ±
20
1
fx 2x
2
4 5x3 4x2 5x 2 Based on the graph, try x 2 and x
2 2 5 4 5 2
2
4
1
2
2
4
1
2
0
1
2 .
1
2
2
2
1
1
0
2
0
2
1
1
0
The zeros of are
The zeros of are x 2, x 1
2x x ±i.
fx
2, and x ±i.
2 2 2x2 1
64
48
16
16
16
The zeros of gx are x 2 and x 1 ± 3i.
0
32
32
0
3
± 2
3i.

79.
Let fx x4 gx 5x
2.
5 10x 5xx4 2
81.
83.
Sign variations: 0, positive zeros: 0
fx x 4 2
Sign variations: 0, negative zeros: 0
hx 3x 4 2x 2 1
Sign variations: 0, positive zeros: 0
hx 3x 4 2x 2 1
Sign variations: 0, negative zeros: 0
gx 2x 3 3x 2 3
Sign variations: 1, positive zeros: 1
gx 2x 3 3x 2 3
Sign variations: 0, negative zeros: 0
85. fx 5x3 x2 x 5
Sign variations: 3, positive zeros: 3 or 1
fx 5x 3 x 2 x 5
Sign variations: 0, negative zeros: 0
1 is a lower bound.
80.
82.
84.
86.
Section 2.5 Zeros of Polynomial Functions 199
hx 4x 2 8x 3
Sign variations: 2, positive zeros: 2 or 0
hx 4x 2 8x 3
Sign variations: 0, negative zeros: 0
hx 2x 4 3x 2
Sign variations: 2, positive zeros: 2 or 0
hx 2x 4 3x 2
Sign variations: 0, negative zeros: 0
f x 4x 3 3x 2 2x 1
Sign variations: 3, positive zeros: 3 or 1
f x 4x 3 3x 2 2x 1
Sign variations: 0, negative zeros: 0
f x 3x 3 2x 2 x 3
Sign variations: 0, positive zeros: 0
f x 3x 3 2x 2 x 3
Sign variations: 3, negative zeros: 3 or 1
87. fx x
(a) 4 1 4 0 0 15
4 0 0 0
1 0 0 0 15
4 is an upper bound.
(b) 1 1 4 0 0 15
1 5 5 5
1 5 5 5 20
4 4x3 15 88. fx 2x
(a) 4 2 3 12 8
8 20 32
2 5 8 40
4 is an upper bound.
(b) 3 2 3 12 8
6 27 45
2 9 15 37
3 3x 2 12x 8
3 is a lower bound.
3 is a lower bound.
89. fx x
(a) 5 1 4 0 16 16
5 5 25 205
1 1 5 41 189
5 is an upper bound.
(b) 3 1 4 0 16 16
3 21 63 141
1 7 21 47 125
4 4x3 16x 16 90. f x 2x
(a) 3 2 0 0
6 18
2 6 18
3 is an upper bound.
(b) 4 2 0
8
2 8
4 8x 3
3 is a lower bound.
8
54
46
0
32
32
3
138
141
8
128
136
3
544
547

200 Chapter 2 Polynomial and Rational Functions
91.
95.
fx 4x 3 3x 1
Possible rational zeros:
1 4
4
0
4
4
3
4
1
1
1
0
4x 3 3x 1 x 14x 2 4x 1
x 12x 1 2
Thus, the zeros are 1 and 1
2.
93. fy 4y
Possible rational zeros:
3 3y2 8y 6
97.
3
4
4
4
3
3
0
8
0
8
Px x 4 25
4 x2 9
1
44x 4 25x 2 36
1
44x 2 9x 2 4
±1, ± 1 1
2 , ± 4
6
6
0
4y 3 3y 2 8y 6 y 3
44y 2 8
Thus, the only real zero is 3
4 .
y 3
44y 2 2
4y 3y 2 2
1
42x 32x 3x 2x 2
The rational zeros are ± and ±2.
3
2
±1, ±2, ±3, ±6, ± 1 3 1 3
2 , ± 2 , ± 4 , ± 4
f x x 3 1
4 x2 x 1
4 98.
1
44x 3 x 2 4x 1
1
4x 2 4x 1 14x 1
1
44x 1x 2 1
1
44x 1x 1x 1
The rational zeros are 1
4 and ±1.
92.
Possible rational zeros:
3
2
f z 12z 3 4z 2 27z 9
12
12
4
18
14
f z 2z 3
26z 2 7z 3
2z 33z 12z 3
Real zeros: 3
2
27
21
6
1 3
, 3 , 2
± 1
±1, ±3, ±9, ±
3 9 1 1
4 , ± 4 , ± 4 , ± 6 , ± 12
1 3 9 1
2 , ± 2 , ± 2 , ± 3 ,
9
9
0
94. gx 3x3 2x 2 15x 10
96.
Possible rational zeros:
2
3
3
3
2
2
0
15
0
15
10
10
gx x 2
33x 2 15 3x 2x 2 5
Thus, the only real zero is 2
3 .
4 2
2
3
8
5
23
20
3
The rational zeros are 3, 1
2 , and 4.
± 5 10
±1, ±2, ±5, ±10, ± 3 , ± 3
1 2
3 , ± 3 ,
0
f x
Possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12,
1
22x 3 3x 2 23x 12
12
12
0
± 1 3
2 , ± 2
f x 1
2x 42x 2 5x 3 1
2x 42x 1x 3
f z 1
66z 3 11z 2 3z 2
Possible rational zeros:
2 6
6
11
12
1
3
2
1
2
2
0
f x 1
6x 26x 2 x 1
1
6x 23x 12x 1
Rational zeros: 2, 1 1
3 , 2
±1, ±2, ± 1 1 2 1
2 , ± 3 , ± 3 , ± 6

99. fx x
Rational zeros: 1 x 1
Irrational zeros: 0
Matches (d).
3 1 x 1x2 x 1
103. (a)
(c)
9
Volume of box
125
100
75
50
25
V
15
x
x
9 − 2x
1 2 3 4 5
Length of sides of
squares removed
The volume is maximum when x 1.82.
x
15 − 2x
The dimensions are: length 15 21.82 11.36
width 9 21.82 5.36
height x 1.82
1.82 cm 5.36 cm 11.36 cm
x
100.
(b)
(d)
Section 2.5 Zeros of Polynomial Functions 201
f x x 3 2
x 3 2x 2 3 2x 3 4
Rational zeros: 0
Irrational zeros:
Matches (a).
1 x 3 2
101. f x x
Rational zeros: 3 x 0, ±1
Irrational zeros: 0
Matches (b).
3 x xx 1x 1 102.
xx
xx 2x 2
Rational zeros: 1 x 0
Irrational zeros: 2 x ±2
Matches (c).
2 f x x
2
3 2x
104. (a) Combined length and width:
(b)
4x y 120 ⇒ y 120 4x
Volume l w h x 2 y
18,000
0 30
0
Dimensions with maximum volume:
20 in. 20 in. 40 in.
x 2 120 4x
4x 2 30 x
(c)
V l w h 15 2x9 2xx
x9 2x15 2x
Since length, width, and height must be positive,
we have 0 < x < for the domain.
9
56 x9 2x15 2x
56 135x 48x 2 4x 3
50 4x 3 48x 2 135x 56
2
1 7
2 ,
The zeros of this polynomial are 2 , and 8.
x cannot equal 8 since it is not in the domain of V.
[The length cannot equal 1 and the width cannot
equal 7.
The product of 817 56 so it
showed up as an extraneous solution.]
Thus, the volume is 56 cubic centimeters when
centimeter or x centimeters.
7
x 2
1
2
15 1
1
13,500 4x 2 30 x
4x 3 120x 2 13,500 0
x 3 30x 2 3375 0
30
15
15
0
225
225
x 15x 2 15x 225 0
3375
3375
Using the Quadratic Formula, x 15,
15 ± 155
.
2
15 155
The value of is not possible because
2
it is negative.
0

202 Chapter 2 Polynomial and Rational Functions
105.
106.
108. (a)
76x 3 4830x 2 2,820,000 0
P 76x 3 4830x 2 320,000, 0 ≤ x ≤ 60
2,500,000 76x 3 4830x 2 320,000
The zeros of this equation are x 46.1, x 38.4, and x 21.0. Since 0 ≤ x ≤ 60, we
disregard x 21.0. The smaller remaining solution is x 38.4. The advertising expense is $384,000.
The zeros of this equation are x 18.0, x 31.5, and
x 42.0. Because 0 ≤ x ≤ 50, disregard x 18.02.
The smaller remaining solution is x 31.5, or an
advertising expense of $315,000.
(b)
P 45x 3 2500x 2 275,000 107. (a) Current bin: V 2 3 4 24 cubic feet
800,000 45x 3 2500x 2 275,000
A 250 x160 x 1.5160250 (c)
60,000 x 2 410x 40,000
x 410 ± 4102 4120,000
21
x must be positive, so
x
0 45x 3 2500x 2 1,075,000
0 9x 3 500x 2 215,000
0 x 2 410x 20,000
410 ± 248,100
2
410 248,100
2
44.05.
60,000
The new length is 250 44.05 294.05 ft and the
new width is 160 44.05 204.05 ft, so the new
dimensions are 204.05 ft 294.05 ft.
109.
C is minimum when 3x3 40x2 C 100
2400x 36000 0.
200 x
, x ≥ 1
x2 x 30
The only real zero is x 40 or 4000 units.
(b)
A 250 2x160 x 60,000
x 570 ± 5702 4220,000
22
x must be positive, so
x
New bin: V 524 120 cubic feet
x 3 9x 2 26x 24 120
x 3 9x 2 26x 96 0
The only real zero of this polynomial is x 2. All the
dimensions should be increased by 2 feet, so the new
bin will have dimensions of 4 feet by 5 feet by 6 feet.
2x 2 570x 20,000 0
570 484,900
4
V 2 x3 x4 x 120
31.6.
The new length is 250 231.6 313.2 ft and the new
width is 160 31.6 191.6 ft, so the new dimensions
are 191.6 ft 313.2 ft.
110. h(t 16t2 48t 6
Let h 64 and solve for t.
64 16t 2 48t 6
16t 2 48t 58 0
48 ± i1408
By the Quadratic Formula we have t .
32
Since the equation yields only imaginary zeros, it is not
possible for the ball to have reached a height of 64 feet.

111.
9,000,000 0.0001x 2 60x 150,000
Thus, 0 0.0001x 2 60x 9,150,000.
x
P R C xp C
x140 0.0001x 80x 150,000
0.0001x 2 60x 150,000
60 ± 60
0.0002
300,000 ± 10,00015i
Since the solutions are both complex, it is not possible
to determine a price p that would yield a profit of
9 million dollars.
113. False. The most nonreal complex zeros it can have is
two and the Linear Factorization Theorem guarantees
that there are 3 linear factors, so one zero must be real.
Section 2.5 Zeros of Polynomial Functions 203
112. (a)
A 0.0167t 3 0.508t 2 5.60t 13.4
(b) 12
The model is a good fit to
the actual data.
7
0
(c) A 8.5 when t 10 which corresponds to the
year 2000.
(d) A 9 when t 11 which corresponds to the
year 2001.
13
(e) Yes. The degree of A is odd and the leading
coefficient is positive, so as x increases, A will increase.
This implies that attendance will continue to grow.
114. False. f does not have real coefficients.
115. gx fx. This function would have the same zeros 116. gx 3f x. This function has the same zeros as f
as fx so r1, r2, and r3 are also zeros of gx.
because it is a vertical stretch of f. The zeros of g are
r1, r2, and r3. 117. gx fx 5. The graph of gx is a horizontal shift 118. Note that x is a zero of g if and only if
of the graph of fx five units to the right so the zeros
is a zero of f. The zeros of g are and
of gx are 5 r1 , 5 r2 , and 5 r3 .
r3 2 .
r1 2 , r2 2 ,
gx f 2x.
2x
119. gx 3 fx. Since gx is a vertical shift of the graph 120. gx f x. Note that x is a zero of g if and only if
of fx, the zeros of gx cannot be determined.
x is a zero of f. The zeros of g are r1, r2, and r3. 121.
fx x 4 4x 2 k
x 2 4 ± 42 41k
21
x ±2 ± 4 k
4 ± 24 k
2
2 ± 4 k
(a) For there to be four distinct real roots, both and must be positive. This occurs
when Thus, some possible k-values are k etc.
1
4 k 2 ± 4 k
0 < k < 4.
k 1, k 2, k 3, , k 2,
(b) For there to be two real roots, each of multiplicity 2, 4 k must equal zero. Thus, k 4.
(c) For there to be two real zeros and two complex zeros, must be positive and
must be negative. This occurs when Thus, some possible k-values are k 1, k 2, k etc.
1
2 4 k
2 4 k
k < 0.
(d) For there to be four complex zeros, 2 ± 4 k must be nonreal. This occurs when k > 4. Some possible
k-values are k 5, k 6, k 7.4, etc.
122. (a) gx f x 2 (b) gx f 2x
No. This function is a horizontal shift of f x.
No. Since x is a zero of g if and only if 2x is a zero of f,
Note that x is a zero of g if and only if x 2 is a
the number of real and complex zeros of g is the same as
zero of f; the number of real and complex zeros is
not affected by a horizontal shift.
the number of real and complex zeros of f.
2
2 ,

204 Chapter 2 Polynomial and Rational Functions
123. Zeros:
f x x 22x 1x 3
8
4
(−2, 0)
−8
2, 1
2 , 3
2x 3 3x 2 11x 6
−4
y
((
1
, 0
2
(3, 0)
4 8 12
x
Any nonzero scalar multiple of f would have the same
three zeros. Let gx af x, a > 0. There are infinitely
many possible functions for f.
125. Answers will vary. Some of the factoring techniques are:
1. Factor out the greatest common factor.
2. Use special product formulas.
127. (a)
a 2 b 2 a ba b
a 2 2ab b 2 a b 2
a 2 2ab b 2 a b 2
a 3 b 3 a ba 2 ab b 2
a 3 b 3 a ba 2 ab b 2
3. Factor by grouping, if possible.
4. Factor general trinomials with binomial factors
by “guess-and-test” or by the grouping method.
5. Use the Rational Zero Test together with synthetic
division to factor a polynomial.
6. Use Descartes’s Rule of Signs to determine the
number of real zeros. Then find any zeros and
use them to factor the polynomial.
7. Find any upper and lower bounds for the real zeros
to eliminate some of the possible rational zeros.
Then test the remaining candidates by synthetic
division and use any zeros to factor the polynomial.
(b)
124.
(−1, 0)
50
10
y
(1, 0) (4, 0)
x
(3, 0) 4 5
126. (a) Zeros of fx: 2, 1, 4
(b) The graph touches the x-axis at x 1
(c) The least possible degree of the function is 4 because
there are at least 4 real zeros (1 is repeated) and a
function can have at most the number of real zeros
equal to the degree of the function. The degree
cannot be odd by the definition of multiplicity.
(d) The leading coefficient of f is positive. From the
information in the table, you can conclude that the
graph will eventually rise to the left and to the right.
(e) Answers may vary. One possibility is:
(f)
fx x 1 2 x 2x 4
x 1 2 x 2x 4
x 2 2x 1x 2 2x 8
x 4 4x 3 3x 2 14x 8
y
(−2, 0)
2 (1, 0) (4, 0)
−3 −1
−4
−6
−8
−10
2 3 5
x2 2ax a2 b2 x a2 bi2 fx x bix bi x
fx x a bix a bi
x a bix a bi
2 b 128. (a) f x
cannot have this graph since it also has a zero at
x 0.
(b) gx cannot have this graph since it is a quadratic
function. Its graph is a parabola.
(c) hx is the correct function. It has two real zeros,
x 2 and x 3.5, and it has a degree of four,
needed to yield three turning points.
(d) k x cannot have this graph since it also has a zero at
x 1. In addition, since it is only of degree three,
it would have at most two turning points.
x

129. 3 6i 8 3i 3 6i 8 3i 11 9i
131. 6 2i1 7i 6 42i 2i 14i 2 20 40i
Section 2.6 Rational Functions
Section 2.6 Rational Functions 205
■ You should know the following basic facts about rational functions.
(a) A function of the form where and are polynomials, is called
a rational function.
(b) The domain of a rational function is the set of all real numbers except those which make the denominator zero.
(c) If is in reduced form, and a is a value such that then the line is a
vertical asymptote of the graph of f.
(d) The line is a horizontal asymptote of the graph of f if or
(e) Let fx where Nx and Dx have no common factors.
1. If n < m, then the x-axis y 0 is a horizontal asymptote.
Nx
Dx anxn an1xn1 . . . a1x a0 bmxm bm1xm1 . . . f x NxDx, Dx 0, Nx Dx
f x NxDx
Da 0,
x a
fx→ or fx→ as x→a.
y b
fx → b as x → x → .
b1x b0 2. If then y is a horizontal asymptote.
an n m,
b m
3. If n > m, then there are no horizontal asymptotes.
Vocabulary Check
130. 12 5i 16i 12 11i
132. 9 5i9 5i 81 25i 2 81 25 106
133. gx fx 2 134. gx f x 2 135. gx 2fx
136.
4
3
2
1
−1
−2
y
(0, 0)
(2, 2)
(4, 2)
(6, 4)
2 3 4 5 6
Horizontal shift two units
to the right
gx f x 137.
(−4, 4)
(−2, 2)
−4 −3 −2 −1
−1
−2
Reflection in the y-axis
4
3
1
y
(0, 2)
1
x
(2, 0)
x
2
3
2
(0, 0)
−2 −1 1 2 3 4
−2
(−2, −2)
−3
y
Vertical shift two units downward
1. rational functions 2. vertical asymptote 3. horizontal asymptote 4. slant asymptote
(2, 0)
(4, 2)
gx f2x 138.
(−1, 0)
4
3
(0, 2)
y
Horizontal shrink each x-value is
multiplied by 1
2
(1, 2)
(2, 4)
−2 −1
1 2
x
x
10
8
6
(0, 4)
(−2, 0)
−2
y
(2, 4)
(4, 8)
2 4 6 8
Vertical stretch (each y-value is
multiplied by 2)
gx f 1
2 x
10
8
6
4
(0, 2)
(−4, 0)
−2
y
(4, 2)
(8, 4)
2 4 6 8
Horizontal stretch each
x-value
is multiplied by 2
x
x

206 Chapter 2 Polynomial and Rational Functions
1. fx
(a)
1
x 1
x
0.5
0.9
0.99
0.999
3. fx
(a)
3x2
x2 1
x f x x
0.5
0.9
0.99
0.999
1
12.79
147.8
1498
4. f x
(a)
4x
x2 1
x f x x
0.5
2.66
0.9 18.95
0.99 199
0.999 1999
1.5
1.1
1.01
1.001
1.5 4.8
f x x
5.4
17.29
152.3
1502
1.1 20.95
1.01 201
1.001 2001
5. fx
Domain: all real numbers except x 0
1
x2 Vertical asymptote:
f x x
2
10
100
1000
2. f x
(a)
5x
x 1
x f x x
0.5 5
0.9 45
0.99 495
0.999 4995
x 0
1.5
1.1
1.01
1.001
1.5 15
1.1 55
Horizontal asymptote: y 0
Degree of Nx < degree of Dx
2
10
100
1000
1.01 505
1.001 5005
f x x
f x x
f x x
5
5
10
100
1000 0.001
10
5
10
100
f x
0.25
0.1
0.01
5 6.25
10
100
f x
5.55
5.05
1000 5.005
1000 3
f x
0.833
0.40
100 0.04
1000 0.004
f x
3.125
3.03
3.0003
(b) The zero of the denominator is x 1, so x 1 is a
vertical asymptote. The degree of the numerator is less
than the degree of the denominator so the x-axis, or
y 0, is a horizontal asymptote.
(c) The domain is all real numbers except x 1.
(b) The zero of the denominator is so is a
vertical asymptote. The degree of the numerator is equal
to the degree of the denominator, so the line y
is a horizontal asymptote.
5
x 1, x 1
1 5
(c) The domain is all real numbers except x 1.
(b) The zeros of the denominator are so both
are vertical asymptotes. Since
the degree of the numerator equals the degree
of the denominator, y is a horizontal
asymptote.
3
x ±1
x 1 and x 1
1 3
(c) The domain is all real numbers except x ±1.
(b) The zeros of the denominator are x ±1 so both
x 1 and x 1 are vertical asymptotes.
Because the degree of the numerator is less than
the degree of the denominator, the x-axis or y 0
is a horizontal asymptote.
(c) The domain is all real numbers except x ±1.
4
6. fx
x 2
Domain: all real numbers except x 2
3
Vertical asymptote:
x 2
Horizontal asymptote: y 0
Degree of Nx < degree of Dx

2 x x 2
7. fx
2 x x 2
Domain: all real numbers except x 2
11.
Vertical asymptote:
Horizontal asymptote: y 1
Degree of Nx degree of Dx
fx
Domain: All real numbers. The denominator has no real
zeros. [Try the Quadratic Formula on the
denominator.]
3x2 1
x2 x 9
Vertical asymptote: None
Horizontal asymptote: y 3
Degree of Nx degree of Dx
13. fx
Vertical asymptote: y 3
2
x 3
Horizontal asymptote: y 0
Matches graph (d).
x 2
9. fx
Domain: all real numbers except x ±1
x3
x2 1
Vertical asymptotes:
x ±1
Horizontal asymptote: None
Degree of Nx > degree of Dx
16.
x 2
f x
x 4
Vertical asymptote: x 4
Horizontal asymptote:
Matches graph (b).
y 1
12.
14. fx
Vertical asymptote: x 5
1
x 5
Horizontal asymptote:
Matches graph (a).
8.
fx
Section 2.6 Rational Functions 207
Domain: all real numbers except
Vertical asymptote:
Horizontal asymptote: y
Degree
of Nx degree of Dx
5
x
2
1
x
2
1
2
fx
Domain: All real numbers. The denominator has
no real zeros. [Try the Quadratic Formula
on the denominator.]
3x2 x 5
x2 1
Vertical asymptote: None
Horizontal asymptote: y 3
Degree of Nx degree of Dx
y 0
1 5x 5x 1
1 2x 2x 1
10. fx
Domain: all real numbers except x 1
2x2
x 1
Vertical asymptote:
17. gx
The only zero of gx is x 1.
x 1 makes gx undefined.
x2 1 x 1x 1
x 1 x 1
x 1
Horizontal asymptote: None
Degree of Nx > degree of Dx
15.
x 1
fx
x 4
Vertical asymptote: x 4
18.
Horizontal asymptote:
Matches graph (c).
hx 2 5
x 2 2
2 5
x 2 2
2x 2 2 5
x2 5
2
2
y 1
0 2 5
x 2 2
No real solution, hx has no real
zeros.

208 Chapter 2 Polynomial and Rational Functions
19.
23.
25.
f x 1 3
x 3
1 3
0
x 3
x 3 3
f x
Domain: all real numbers x except x 1 and x 3
x2 1
x2 x 1x 1 x 1
, x 1 24.
2x 3 x 1x 3 x 3
Horizontal asymptote:
y 1
Degree of Nx degree of Dx
Vertical asymptote: x 3
Since
x 1 is a common factor of Nx and Dx, x 1
is not a vertical asymptote of f x.
f x x2 3x 4
2x 2 x 1
1 3
x 3
x 6 is a zero of f x.
x 1x 4
2x 1x 1
x 4
, x 1
2x 1
Domain: all real numbers except and
Horizontal asymptote:
Vertical asymptote: x Since
x 1 is a common
factor of Nx and Dx, x 1 is not a vertical asymptote
of f x.
1
y
Degree of Nx degree of Dx
2
1
x x 1
2
1
x
2
20.
26.
gx x3 8
x 2 1
x3 8
x2 0
1
x 3 8 0
x 3 8
x 2 is a real zero of gx.
21.
x 4
f x
x
Domain: all real numbers x except x ±4
Horizontal asymptote: y 0
Degree of Nx < degree of Dx
Vertical asymptote: x 4 Since
x 4 is a common
factor of Nx and Dx, x 4 is not a vertical asymptote
of f x.
2 1
, x 4
16 x 4
22.
x 3
f x
x
Domain: all real numbers x except x ±3
The degree of the numerator is less than the degree of the
denominator, so the graph has the line y 0 as a
horizontal asymptote.
Vertical asymptote: x 3 Since
x 3 is a common
factor of Nx and Dx, x 3 is not a vertical asymptote
of f x.
2 9
x 3 1
, x 3
x 3x 3 x 3
f x x2 4
x 2 3x 2
x 2x 2
x 2x 1
Domain: all real numbers x except x 1 and x 2
Horizontal asymptote:
x 2
, x 2
x 1
y 1
Degree of Nx degree of Dx
Vertical asymptote: x 1 Since
x 2 is a common
factor of Nx and Dx, x 2 is not a vertical asymptote
of f x.
f x 6x2 11x 3
6x 2 7x 3
x 2
2x 33x 1
2x 33x 1
Domain: all real numbers x except x or
Horizontal asymptote: y 1
3
2
Degree of Nx degree of Dx
3x 1 3
, x
3x 1 2
x 1
3
Vertical asymptote: Since is a common
factor of and x is not a vertical asymptote
of f x.
3
x 2x 3
Nx Dx, 2
1
3

27. fx
(a) Domain: all real numbers x except x 2
1
x 2
(b) y-intercept:
(c) Vertical asymptote: x 2
Horizontal asymptote: y 0
(d)
29. hx
(a) Domain: all real numbers x except x 2
(b) y-intercept: 0, 1 2
(c) Vertical asymptote: x 2
Horizontal asymptote: y 0
1
x 2
(d)
x 0
y 1 1
4 3 1
1
1
2
1
0, 2
x 4 3 1 0 1
y 1 1
1
2
−3 −1
−4 −3 −1
(( 0, − 1
2
2
1
−1
−2
2
1
−1
−2
y
y
(( 0, 1
Note: This is the graph of fx
(Exercise 27) reflected about the x-axis.
1
x 2
1
2
2
x
x
2
1
3
Section 2.6 Rational Functions 209
28. f x
(a) Domain: all real numbers x except x 3
(b)
1
y- intercept: 0, 3
(c) Vertical asymptote: x 3
Horizontal asymptote: y 0
1
x 3
(d)
30. gx
(a) Domain: all real numbers x except x 3
1 1
3 x x 3
(b) y-intercept:
(c) Vertical asymptote: x 3
Horizontal asymptote: y 0
(d)
3
2
1
−1
−2
−3
3
2
1
−1
−2
−3
x 0 1 2 4 5 6
y
y
(( 0,
(( 0, 1
1
3
y 1 1
1
−
3
3
1
0, 3
x 0 1 2 4 5 6
y 1 1 1
1 1
3
2 4 5 6
2
1
2
1 2 4
Note: This is the graph of fx
(Exercise 28) reflected about the x-axis.
1
x 3
x
x
1
2
1
2
1
3
3

210 Chapter 2 Polynomial and Rational Functions
31.
5 2x 2x 5
Cx
1 x x 1
(a) Domain: all real numbers x except x 1
(b) x-intercept:
33. fx
(a) Domain: all real numbers x
x2
x2 9
(b) Intercept:
(c) Horizontal asymptote: y 1
(d)
y-intercept:
(c) Vertical asymptote: x 1
Horizontal asymptote: y 2
(d)
x
y
((
5
− , 0
2
−6 −4 2 4
±1
1
10
3
2
−2 −1 1 2
−1
0, 0
y
5, 0 2
0, 5
x 4 3 2 0 1 2
Cx
y
±2
6
(0, 5)
4
13
(0, 0)
1
2
1 1 5 3
±3
1
2
x
x
7
2
1 3x 3x 1
32. Px
1 x x 1
(a) Domain: all real numbers x except x 1
34.
(b) x-intercept:
1 2t 2t 1
ft
t t
(a) Domain: all real numbers t except t 0
(b) t-intercept:
(c) Vertical asymptote: t 0
Horizontal asymptote: y 2
(d)
y-intercept:
(c) Vertical asymptote: x 1
Horizontal asymptote: y 3
(d)
6
5
4
y
(0, 1) 1
, 0
3
−2 −1 2 3 4
−2 −1 1 2
−1
−3
y
1
, 0 2
t 2 1 1 2
y 3 0 1 3
2
5
2
1
, 0 3
0, 1
x 1 0 2 3
y 2 1 5 4
( )
1 ( , 0)
2
1
x
t
2

35. gs
(a) Domain: all real numbers s
s
s2 1
37.
(b) Intercept:
(c) Horizontal asymptote:
(d)
hx x2 5x 4
x 2 4
(a) Domain: all real numbers x except x ±2
(b) x- intercepts: 1, 0, 4, 0
y-intercept:
0, 1
(c) Vertical asymptotes: x 2, x 2
Horizontal asymptote: y 1
(d)
x
y
4
10
3
2
1
(0, 0)
−1
−2
0, 0
6
4
2
y
−6 −4
(4, 0) 6
y
3
28
5
1 2
x 1x 4
x 2x 2
(1, 0)
1
10
3
y 0
s 2 1 0 1 2
gs
2
5
1
2
0
1
2
s
x
2
5
0 1 3 4
0 0
2
1
5
38.
gx x2 2x 8
x 2 9
Section 2.6 Rational Functions 211
36. fx
(a) Domain: all real numbers x except x 2
(b)
1
y- intercept: 0, 4
(c) Vertical asymptote: x 2
Horizontal asymptote: y 0
1
x 22 (d)
(a) Domain: all real numbers x except x ±3
8
(b) y- intercept: 0, 9
x-intercepts:
4, 0, 2, 0
(c) Vertical asymptotes: x ±3
Horizontal asymptote: y 1
(d)
x
y
−1
−2
−3
−4
x
y
0 1 3 4
1
4
y
(( 0, − 1
4
5
27
16
1
2
4
9
1 3
6
4
2
4
16
7
1
y
(0, 0.88)
(4, 0)
−6 −4
(−2, 0) −2
−4
−6
2 4 6
3
2
x 4x 2
x 3x 3
2
4
x
x
0 2 4 5
8
9
5
2
4
8
5
0 0
1
7
16
4
7
2
9
1
4

212 Chapter 2 Polynomial and Rational Functions
39.
f x
(a) Domain: all real numbers x except x 2, x ±1
2x2 5x 3
x3 2x2 x 2
2x 1x 3
x 2x 1x 1
(b) x-intercepts:
y-intercept:
(c) Vertical asymptotes: x 2,
x 1 and x 1
Horizontal asymptotes: y 0
(d)
1
, 0
2
9
6
3
(3, 0)
−4 −3 3 4
(( −
((
0, − 3
2
41. f x
(a) Domain: all real numbers x except x 3 and x 2
(b) Intercept: 0, 0
(c) Vertical asymptote: x 2
Horizontal asymptote: y 1
(d)
x2 3x
x2 x 6
xx 3 x
, x 3
x 3x 2 x 2
42.
5x 4
f x
x
(a) Domain: all real numbers x except x 4 or x 3
(b) y- intercept: 0, 5 3
x-intercept:
none
(c) Vertical asymptote: x 3
Horizontal asymptote: y 0
(d)
2 x 12
5x 4 5
, x 4
x 4x 3 x 3
x 1 0 1 3 4
y
1
3 0 1 3 2
x 2 0 2 5 7
6
4
2
−6 −4 −2
4 6
(0, 0)
−4
−6
y
y
0, 3
1,
0 , 3, 0
2
2
x 3 2 0 1.5 3 4
f x
3
4
5
4
3
2
x
x
48
5
0
3
10
40.
x
f x
(a) Domain: all real numbers x except x 1, x 2,
or x 3
(b) x-intercepts: 1, 0, 2, 0
y-intercept: 0, 1 3
(c) Vertical asymptotes: x 2, x 1, x 3
Horizontal asymptote: y 0
2 x 2
x3 2x2 5x 6
x 1x 2
x 1x 2x 3
(d)
x 4
3 1 0 2 4
y 0 0
( −1,
0)
((
1
0, −
3
y
6
4
2
(0, −1.66)
9
35
1
−4
−6
y
4
3
2
1
−2
−3
−4
−5
y
5
12
(2, 0)
5
3
2 4 5
5
2 4 6 8
x
x
5
2
1
3
5
4
5
9

43.
45.
f x 2x2 5x 2
2x 2 x 6
(a) Domain: all real numbers except and
x 3
x x 2
2
(b) x- intercept:
(d)
f t t 2 1
t 1
(a) Domain: all real numbers t except t 1
(b) t- intercept: 1, 0
y-intercept:
0, 1
(c) Vertical asymptote: none
Horizontal asymptote: none
(d)
y-intercept:
(c) Vertical asymptote: x
Horizontal asymptote: y 1
3
2
x
y
2x 1x 2
2x 3x 2
−5 −4 −3 −2 1 2 3
, 0
1)0,
2 )
−
3 )
)
3
7
3
3
2
1
t 1t 1
t 1
−3 −2 −1 1 2 3
−1
−2
−3
(0, −1)
y
1
, 0 2
0, 1 3
4
3
2
1
2
5
y
(1, 0)
2x 1
, x 2
2x 3
1
3
x
t
0 1
1
3
1
5
t 1, t 1 46.
Section 2.6 Rational Functions 213
f x
(a) Domain: all real numbers x 4
x2 16 x 4x 4
x 4, x 4
x 4 x 4
(b) y- intercept: 0, 4
x-intercept:
4, 0
(c) Vertical asymptote: none
Horizontal asymptote: none
t 3 2 0 1
(d)
x 6 4 0 5
y 4 3 1 0
y 2 0 4 9
44.
f x 3x2 8x 4
2x 2 3x 2
(a) Domain: all real numbers x except x 2 or
(b) y- intercept: 0, 2
x-intercept:
(c) Vertical asymptote:
Horizontal asymptote: y
(d)
3
x
2
1
2
x
3 1 0
2
3
y
−4 −3 −2 −1 2 3 4
, 0
3
(−4, 0)
x 23x 2
x 22x 1
11
5
1
10
8
6
4
2
y
(0, 4)
−6 −2 2 4 6
−2
y
2
, 0 3
((
(0, −2)
3x 2
, x 2
2x 1
5 2 0 1
x
x
3
x 1
2

214 Chapter 2 Polynomial and Rational Functions
47. fx
(a) Domain of f : all real numbers x except x 1
Domain of g: all real numbers x
(b) Because x 1 is a factor of both the numerator
and the denominator of f, x 1 is not a vertical
asymptote. f has no vertical asymptotes.
x2 1
, gx x 1
x 1
(c)
(d)
49.
x 2
fx
x
(a) Domain of f : all real numbers x except x 0 and
x 2
2 1
, gx
2x x
Domain of g: all real numbers x except x 0
(b) Because x 2 is a factor of both the numerator and
the denominator of f, x 2 is not a vertical asymptote.
The only vertical asymptote of f is x 0.
(c)
(d)
−4
−3
x 3 2 1.5 1 0.5 0 1
f x
gx
x 0.5 0 0.5 1 1.5 2 3
f x
gx
4
4
(e) Because there are only a finite number of pixels,
the utility may not attempt to evaluate the function
where it does not exist.
2
2
2
−2
1
−3
3
3
2.5
2.5
2
Undef. 2 1 Undef.
Undef. 2 1
3
Undef. 1.5 1 0
2
(e) Because there are only a finite number of pixels,
the utility may not attempt to evaluate the function
where it does not exist.
2
3
2
3
1.5
1
2
1
1
3
1
3
0
48. fx
(a) Domain of f : all real numbers x except 0 and 2
x2x 2
x2 , gx x
2x
2x 6
50. fx
x
(a) Domain of f : all real numbers x except 3 and 4
2 2
, gx
7x 12 x 4
Domain of g: all real numbers x except 4
(b) Since x 3 is a factor of both the numerator and the
denominator of f, x 3 is not a vertical asymptote
of f. Thus, f has x 4 as its only vertical asymptote.
(c)
(d)
Domain of g: all real numbers x
(b) Since x is a factor of both the numerator
and the denominator of f, neither x 0 nor x 2
is a vertical asymptote of f. Thus, f has no
vertical asymptotes.
2 2x
(c)
(d)
−2
−1
x 1 0 1 1.5 2 2.5 3
f x
(e) Because there are only a finite number of pixels,
the utility may not attempt to evaluate the function
where it does not exist.
x 0 1 2 3 4 5 6
f x
3
−3
2
−2
1 Undef. 1 1.5 Undef. 2.5 3
g(x) 1 0 1 1.5 2 2.5 3
1
2
1
2
2
3
8
1 Undef. Undef. 2 1
g(x) 1 2 Undef. 2 1
2
3
4
(e) Because there are only a finite number of pixels, the
utility may not attempt to evaluate the function where
it does not exist.

51.
53.
55.
hx
(a) Domain: all real numbers x except x 0
x2 4 4
x
x x
(b) Intercepts:
(c) Vertical asymptote: x 0
Slant asymptote: y x
(d)
(−2, 0)
fx
(a) Domain: all real numbers x except x 0
(b) No intercepts
(c) Vertical asymptote: x 0
Slant asymptote: y 2x
(d)
2x2 1
2x
x
1
x
x 4 2 2 4 6
f x
2
33
4
6
4
2
y = 2x
−6 −4 −2 2 4 6
−6
2, 0, 2, 0
−6 −4
(2, 0) 6
−2
−4
−6
y
y
y = x
9
2
9
2
x
x
33
4
73
6
52.
54.
(b) No intercepts
fx
Section 2.6 Rational Functions 215
(c) Vertical asymptote: x 0
Slant asymptote: y x
(d)
x 3 1 1 3
x 3 2 1 1 2 3
y
5
3
3
3
gx
(a) Domain: all real numbers x except x 0
x2 1 1
x
x x
(b) No intercepts
(c) Vertical asymptote: x 0
Slant asymptote: y x
5
3
(d)
gx
(a) Domain: all real numbers x except x 0
x2 5 5
x
x x
(a) Domain: all real numbers x except x 0
(b) x-intercepts: 1, 0, 1, 0
(c) Vertical asymptote: x 0
Slant asymptote: y x
(d)
y
−6 −4 −2 2 4 6
−2
−4
x
1 x2
x
f x
14
3
6
35
6
y = −x
8
6
4
2
(−1, 0) (1, 0)
−8 −6 −4 −2 4 6 8
−4
−6
−8
6
4
2
x 1
x
y
y
9
2
4
15
4
6
y = x
2
3
2
x 4
2 2 4 6
gx
17
4
5
2
5
2
17
4
37
6
x
6
x
9
2
2 4 6
3
2
14
3
6
4
2
15
4
35
6
−6 −4 −2 2 4 6
−6
y
y = x
x

216 Chapter 2 Polynomial and Rational Functions
56. hx
(a) Domain: all real numbers x except x 1
x2
1
x 1
x 1 x 1
58.
(b) Intercept:
(c) Vertical asymptote: x 1
Slant asymptote: y x 1
(d)
x 4 2 2 4 6
fx x2
3x 1
(a) Domain: all real numbers x except
(b) Intercept:
(c) Vertical asymptote:
Slant asymptote: y 1
x
1
x
3 9
1
3
(d)
hx
x
y
1 1
y = x −
3 9
−1
8
6
4
2
(0, 0)
−4 2 4 6 8
−2
−4
y
3
9
8
1
2
3
1
3
0, 0
16
5
1 1
x
3 9
0, 0
y
y = x + 1
2
4
5
(0, 0)
1
3
4
3
2
3
1
1
1
2
4
3
4
x
1
93x 1
x
16
3
1
2
1
2
36
5
x 1
3
0 2
0
4
7
57.
ft
(a) Domain: all real numbers t except t 5
(b) Intercept: 0, 1 5
(c) Vertical asymptote: t 5
Slant asymptote: y t 5
t 2 1
26
t 5
t 5 t 5
(d)
t
y
59. fx
(a) Domain: all real numbers x except x ±1
(b) Intercept: 0, 0
(c) Vertical asymptotes: x ±1
Slant asymptote: y x
(d)
x3
x2 x
x
1 x2 1
x 4 2 0 2 4
f x
0
25 37 1
7 6 4 3
17 5
y = 5 − t
(0, −0.2)
−20 −15 −10 −5
10
64
15
−6 −4 −2 2 4 6
y
2
25
20
15
5
y
8
3
(0, 0)
0
y = x
t
x
8
3
64
15
5

60. gx
(a) Domain: all real numbers x except x ±2
x3
2x2 1 4x
x
8 2 2x2 8
62.
(b) Intercept:
(c) Vertical asymptotes:
(d)
fx
(a) Domain: all real numbers x except x 2
5
(b) y-intercept: 0, 2
(c) Vertical asymptote: x 2
Slant asymptote: y 2x 1
2x2 5x 5
2x 1
x 2
3
x 2
(d)
Slant asymptote:
x
gx
x
y
6
107
8
15
12
9
6
3
3
38
5
−9 −6 −3 3 6 9 12 15
(( 0, − 5
(0, 0)
2
−9
y
8
6
4
0, 0
6
27
8
−8 −6 −4 4 6 8
y
y = 2x − 1
y 1
2 x
4
8
3
y =
1
x
2
x ±2
1
1
6
x
1 3 6 7
2
x
8
1 4 6
1
6
47
4
8
3
68
5
27
8
61.
63.
Section 2.6 Rational Functions 217
fx
(a) Domain: all real numbers x except x 1
x2 x 1 1
x
x 1 x 1
(b) y-intercept:
(c) Vertical asymptote: x 1
Slant asymptote: y x
(d)
f x 2x3 x 2 2x 1
x 2 3x 2
(a) Domain: all real numbers except and
(b) intercept:
intercepts:
(c) Vertical asymptote: x 2
Slant asymptote: y 2x 7
1
, 0 , 1, 0
2 x-
2x 7
x x 1 x 2
1
y- 0, 2
15
, x 1
x 2
(d)
2x 1x 1x 1
x 1x 2
2x 1x 1
, x 1
x 2
2x2 3x 1
x 2
x
y
8
6
4
2
(0, −1)
−4 −2 2 4 6 8
−4
y
4
45
2
0, 1
x 4
2 0 2 4
f x
21
5
18
3
28
12 (0, 0.5)
6 (1, 0)
x
−6 −5 −4 −3 −1
3
−12 (0.5, 0)
−18
−24
−30
−36
y = 2x − 7
y
7
3
y = x
1
3
2
x
3
13
3
0 1
1
20 2 0

218 Chapter 2 Polynomial and Rational Functions
64.
66.
68.
f x 2x3 x 2 8x 4
x 2 3x 2
(a) Domain: all real numbers except or
(b) intercept:
intercepts: 2, 0,
(c) Vertical asymptote: x 1
Slant asymptote: y 2x 7
1
x x 1 x 2
y- 0, 2
x-
, 0 2
(d)
fx
Domain: all real numbers x except x 1
2x2 x
1
2x 1
x 1 x 1
Vertical asymptote:
x 1
Slant asymptote: y 2x 1 −12
Line: y 2x 1
12 2x x2 1 2
hx x 1
24 x 2 4 x
Domain: all real numbers x except x 4
Vertical asymptote:
Slant asymptote:
Line: y 1
x 1
2
70. (a) x-intercept:
(b)
2x 7 9
, x 2
x 1
x
y
0 2x
x 3
0 2x
0 x
3
5
4
(−2, 0)
30
24
18
12
−6 −2
(0, −2)
2 4 6
y
2
((
− , 0
1
2
1
1
2
y = 2x + 7
x 4
y 1
2 x 1 −16
x
x 2x 22x 1
x 2x 1
1
2
0 3 4
0 2 10 28 18
6
−10
3
2
10
−6
35
2
12
8
67.
gx
Domain: all real numbers x except x 0
1 3x2 x3 x2 1
1
3 x x 3
x2 x2 Vertical asymptote:
Slant asymptote:
Line: y x 3
(b)
0
0 x 1
1 x
x 1
x 3
x 0
y x 3
69.
x 1
y
x 3
(a) x-intercept: 1, 0
0, 0 71. y
(a) x-intercepts: ±1, 0 (b)
1
x
x
65.
fx
Domain: all real numbers x except x 3
x2 5x 8
x 2
x 3
2
x 3
y-intercept:
Vertical asymptote:
Slant asymptote:
Line:
−14
8
0, 3
y x 2
8
−8
x 3
y x 2
10
−12
x 2 1
x ±1
12
−4
x 1
0
x
1
x
x
12

72. (a) x-intercepts:
(b) 0 x 3 2
1, 0, 2, 0 73. C
x
(a) 2,000
255p
, 0 ≤ p < 100
100 p
74. C
(a) 300,000
25,000p
, 0 ≤ p < 100
100 p
75.
205 3t
N , t ≥ 0
1 0.04t
(a) N5 333 deer
(b)
76. (a)
0 x 2 3x 2
0 x 1x 2
x 1, x 2
0
0
C
The cost would be $4411.76.
25,00015
4411.76
100 15
C 25,00050
100 50
100
25,000
The cost would be $25,000.
C
The cost would be $225,000.
(c) C → as x → 100. No. The model is undefined for
p 100.
25,00090
225,000
100 90
0.2550 0.75x C50 x (c)
C
C
(b) Domain:
12.50 0.75x
50 x
4
4
50 3x 3x 50
450 x 4x 50
x ≥ 0 and x ≤ 1000 50
Thus, 0 ≤ x ≤ 950. Using interval notation, the
domain is 0, 950.
(b)
0
0
Section 2.6 Rational Functions 219
C10 25510
28.33 million dollars
100 10
C40 25540
170 million dollars
100 40
C75 25575
765 million dollars
100 75
(c) C → as x → 100. No, it would not be possible
to remove 100% of the pollutants.
N10 500 deer
N25 800 deer
(b) The herd is limited by the horizontal asymptote:
N 60
1500 deer
0.04
1.0
0.8
0.6
0.4
0.2
C
100
200 400 600 800 1000
(d) As the tank is filled, the concentration increases
more slowly. It approaches the horizontal asymptote
of C 3
4 0.75.
x

220 Chapter 2 Polynomial and Rational Functions
77. (a) A xy and
(c)
78. A xy and
x 3y 2 64
Thus,
x 4y 2 30
200
4
0
y 2 64
x 3
y 2 64 2x 58
x 3 x 3
2x 58
A xy x x 3
2xx 29
, x > 3.
x 3
By graphing the area function, we see that A is minimum when x 12.8 inches and y 8.5 inches.
79. (a) Let time from Akron to Columbus and
from Columbus back to Akron.
time
xt1 100 ⇒ t1 100
t1 t2
(b) Vertical asymptote: x 25
Horizontal asymptote: y 25
x
(c) 200
yt 2 100 ⇒ t 2 100
y
50t 1 t 2 200
100
x
y 2 30
x 4
The area is minimum when x 11.75 inches and y 5.87 inches.
The area is minimum when x is approximately 12.
t 1 t 2 4
100
y
4
100y 100x 4xy
25y 25x xy
25x xy 25y
25x yx 25
Thus, y 25x
x 25 .
y 2 30 2x 22
x 4 x 4
2x 22 2xx 11
Thus, A xy x x 4 .
x 4
(b) Domain: Since the margins on the left and right are each 2 inches, x > 4. In interval notation, the domain is 4, .
40
x
5 6 7 8 9 10 11 12 13 14 15
y1 (Area) 160 102 84 76 72 70 69.143 69 69.333 70 70.909
200
3
0
(d)
39
25 65
0
x 30 35 40 45 50 55 60
y 150 87.5 66.67 56.25 50 45.83 42.86
(e) Yes. You would expect the average speed for the round
trip to be the average of the average speeds for the two
parts of the trip.
(f) No. At 20 miles per hour you would use more time in
one direction than is required for the round trip at an
average speed of 50 miles per hour.

80. (a)
81. False. Polynomial functions do not have vertical
asymptotes.
83. Vertical asymptote: None ⇒ The denominator is not
zero for any value of x (unless the numerator is also zero
there).
Horizontal asymptote: y 2 ⇒ The degree of the
numerator equals the degree of the denominator.
fx is one possible function. There are many
correct answers.
2x2
x2 1
Section 2.6 Rational Functions 221
82. False. The graph of f x crosses y 0, which
is a horizontal asymptote.
x
x2 1
84. Vertical asymptotes:
are factors of the denominator.
Horizontal asymptotes: The degree of the
numerator is greater than the degree of the denominator.
x
f x
is one possible function. There are
many correct answers.
3
x 2, x 1 ⇒ x 2x 1
None ⇒
x 2x 1
85. x 2 15x 56 x 8x 7 86. 3x 2 23x 36 3x 4x 9
87.
(b) S
The sales in 2008 is estimated to be $763,810,000.
5.816182 130.68
0.004182 763.81
1.00
(c) Probably not. The graph has a horizontal asymptote at
Future sales may exceed this limiting value.
x3 5x2 4x 20 x 5x2 4 88. x3 6x2 2x 12 x2x 6 2x 6
x ≥ 10
3
x 5x 2ix 2i
89. 10 3x ≤ 0 10
3
90. 5 2x > 5x 1
3x ≥ 10
x
5 2x > 5x 5
91.
600
8
0
12 < 4x < 28
3 < x < 7
13
0
1 2 3 4 5 6
4x 2 < 20 92.
−3
20 < 4x 8 < 20 −4 −2 0 2
93. Answers will vary.
4 6
7
8
x
S 5.816
1454 million dollars.
0.004
1
2
7x > 0
x < 0
2x 3 ≥ 5
2x 3 ≥ 10
2x 3 ≤ 10
2x ≤ 13
x ≤ 13
2
or
x 6x 2 2
x 6x 2x 2
2x 3 ≥ 10
−3 −2 −1 0 1 2 3
2x ≥ 7
x ≥ 7
2
−8
13
−
2
−6
−4
−2
0
2
7
2
4
x
x

222 Chapter 2 Polynomial and Rational Functions
Section 2.7 Nonlinear Inequalities
1. x2 3 < 0
2.
■ You should be able to solve inequalities.
(a)
(a)
(a) Find the critical number.
x 3
3 2 3 < ? 0
6 < 0
No, x 3 is not a
solution.
x 2 x 12 ≥ 0
x 5
5 2 5 12 ≥ ? 0
Yes, x 5 is a
solution.
3.
x 2
≥ 3
x 4
(a) x 5
1. Values that make the expression zero
2. Values that make the expression undefined
(b) Test one value in each test interval on the real number line resulting from the critical numbers.
(c) Determine the solution intervals.
Vocabulary Check
1. critical; test intervals 2. zeros; undefined values 3. P R C
5 2
5 4 ≥? 3
7 ≥ 3
Yes, x 5 is a
solution.
8 ≥ 0
(b)
(b)
x 0 (c)
0 2 3 < ? 0
3 < 0
Yes, x 0 is a solution.
x 0 (c)
0 2 0 12 ≥ ? 0
12 ≥ 0
No, x 0 is not a
solution.
(b)
4 2
4 4
6
is undefined.
0
No, x 4 is not a
solution.
≥? x 4
3
(c)
x 3
2
3
2 2 3 < ? 0
3
4 < 0
Yes, x is a solution.
3
2
4 2 4 12 ≥ ? 0
16 4 12 ≥ ? 0
8 ≥ 0
Yes, x 4 is a solution.
(d)
x 4 (d)
x 9
2
92
2
9
2 4 ≥? 3
5
≥ 3
17
No, x is not a
solution.
9
2
(d)
x 5
5 2 3 < ? 0
22 < 0
No, x 5 is not a
solution.
x 3
3 2 3 12 ≥ ? 0
9 3 12 ≥ ? 0
0 ≥ 0
Yes, x 3 is a solution.
x 9
2
9
2
2
9
2 4 ≥? 3
13 ≥ 3
Yes, x is a solution.
9
2

4.
5.
8.
10.
3x2 x2 < 1
4
(a)
x 2
322
2 2 4

224 Chapter 2 Polynomial and Rational Functions
11.
13.
15.
Critical numbers: x 7, x 3
Test intervals: , 7, 7, 3, 3,
Test: Is x 7x 3 < 0?
Interval x-Value Value of
x 7x 3
Conclusion
, 7
7, 3
3,
x 2 4x 4 ≥ 9 14.
x 2 4x 5 ≥ 0
x 5x 1 ≥ 0
Critical numbers: x 5, x 1
Test intervals: , 5, 5, 1, 1,
Test: Is x 5x 1 ≥ 0?
Interval x-Value Value of
x 5x 1
Conclusion
, 5 x 6 17 7 Positive
5, 1 x 0 51 5 Negative
1, x 2 71 7 Positive
Solution set: , 5 1,
−6 −5 −4 −3 −2 −1
0
1
Critical numbers: x 3, x 2
Test intervals: , 3, 3, 2, 2,
Test: Is x 3x 2 < 0?
2
x
Interval x-Value Value of
x 3x 2
Conclusion
, 3
3, 2
2,
x 2 2 < 25
x 2 4x 4 < 25
x 2 4x 21 < 0
x 7x 3 < 0
Solution set:
Solution set:
x 10
x 0
x 5
7, 3
x 2 x < 6 16.
x 2 x 6 < 0
x 3x 2 < 0
x 4
x 0
x 3
3, 2
313 39
73 21
122 24
−7
−8 −6 −4
−2
16 6
32 6
61 6
0 2
−3 −2 −1 0 1 2
3
4
Positive
Negative
Positive
x
Positive
Negative
Positive
x
12.
x 2x 4 ≥ 0
Critical numbers:
Test intervals: , 2 ⇒ x 2x 4 > 0
Solution intervals: , 2 4,
x 1x 7 < 0
Critical numbers:
Test intervals: , 1 ⇒ x 1x 7 > 0
Solution interval: 1, 7
−2
x 2 6x 8 ≥ 0
x 2 6x 9 < 16
x 2 6x 7 < 0
−1
x 3 2 ≥ 1
1 2 3 4 5
0 2 4 6
x 3x 1 > 0
Critical numbers:
x 1, x 7
1, 7 ⇒ x 1x 7 < 0
7, ⇒ x 1x 7 > 0
7
x 2 2x > 3
x 2 2x 3 > 0
8
x
Test intervals: , 3 ⇒ x 3x 1 > 0
Solution intervals: , 3 1,
−4 −3 −2 −1 0 1 2
x 2, x 4
2, 4 ⇒ x 2x 4 < 0
4, ⇒ x 2x 4 > 0
x
x 3, x 1
3, 1 ⇒ x 3x 1 < 0
1, ⇒ x 3x 1 > 0
x

17.
19.
20.
x 2 2x 3 < 0 18.
x 3x 1 < 0
Critical numbers: x 3, x 1
Test intervals: , 3, 3, 1, 1,
Test: Is x 3x 1 < 0?
Interval x-Value Value of
x 3x 1
Conclusion
, 3
3, 1
1,
Solution set:
x 4
x 0
x 2
3, 1
x 2 8x 5 ≥ 0
x 2 8x 5 0
x 2 8x 16 5 16
x 4 2 21
x 4 ±21
15 5
31 3
51 5
−3
x 4 ± 21
−2
−1
0
1
Positive
Negative
Positive
Complete the square.
x
x
Section 2.7 Nonlinear Inequalities 225
x 2 4x 1 > 0
4 ± 16 4
2
Critical numbers:
Test intervals:
Solution intervals:
2 − 5 2 + 5
−4 −2 0 2 4 6 8
Critical numbers:
Test intervals:
Test: Is x2 x 4 ± 21
, 4 21, 4 21, 4 21, 4 21,
8x 5 ≥ 0?
Interval x-Value Value of x Conclusion
, 4 21 x 10 100 80 5 15 Positive
4 21, 4 21 x 0 0 0 5 5 Negative
4 21,
x 2 4 16 5 15 Positive
2 8x 5
2 ± 5
x 2 5, x 2 5
, 2 5 ⇒ x 2 4x 1 > 0
2 5, 2 5 ⇒ x 2 4x 1 < 0
2 5, ⇒ x
, 2 5 2 5,
2 4x 1 > 0
−4 − 21
Solution set: < 4 21 4 21, −10 −8 −6 −4 −2 0 2
2x 2 6x 15 ≤ 0
2 x 2 6x 15 ≥ 0
x 6 ± 62 4215
22
Critical numbers:
Test intervals:
3
2
Solution interval:
x 3
2
3 39
2 2 , ⇒ 2x2 6x 15 < 0
3 39
,
2 2
6 ± 156
4
39 3 39
, x
2 2 2
3 39
,
2 2 ⇒ 2x2 6x 15 < 0
39 3 39
,
2 2 2 ⇒ 2x2 6x 15 > 0
3 39
2 2 ,
6 ± 239
4
3 39
−
2 2
−2 −1
3 39
±
2 2
0
1
2
3
3 39
+
2 2
4
5
x
x
−4 + 21
x

226 Chapter 2 Polynomial and Rational Functions
21.
23.
24.
Critical numbers: x ±1, x 3
Test intervals: , 1, 1, 1, 1, 3, 3,
Test: Is x 1x 1x 3 > 0?
Interval x-Value Value of
x 1x 1x 3
Conclusion
1, 1
1, 3
3,
x 3 3x 2 x 3 > 0 22.
x 2 x 3 1x 3 > 0
, 1
Solution set:
x 3 2x 2 9x 2 ≥ 20
x 3 2x 2 9x 18 ≥ 0
x 2 x 2 9x 2 ≥ 0
x 2x 2 9 ≥ 0
x 2x 3x 3 ≥ 0
Critical numbers: x 2, x ±3
Test intervals: , 3, 3, 2, 2, 3, 3,
Test: Is x 2x 3x 3 ≥ 0?
Interval x-Value Value of x 2x 3x 3 Conclusion
, 3 x 4 617 42 Negative
3, 2 x 0 233 18
Positive
2, 3 x 2.5 0.55.50.5 1.375 Negative
3, x 4 271 14
Positive
Solution set: 3, 2 3,
2x 3 13x 2 8x 46 ≥ 6
2x 3 13x 2 8x 52 ≥ 0
x 2 2x 13 42x 13 ≥ 0
Critical numbers:
Test intervals:
x 2 1x 3 > 0
x 1x 1x 3 > 0
x 2
x 0
x 2
x 4
1, 1 3,
2x 13x 2 4 ≥ 0
113 3
311 3
531 15
x 13
2 , x 2, x 2
, 13
2 ⇒ 2x 3 13x 2 8x 52 < 0
13
2 , 2 ⇒ 2x3 13x 2 8x 52 > 0
2, 2 ⇒ 2x 3 13x 2 8x 52 < 0
2, ⇒ 2x 3 13x 2 8x 52 > 0
Solution interval: 13
2 , 2, 2,
135 15
−1
Negative
Positive
Negative
Positive
0 1 2 3 4
x
13
−
2
x 3 2x 2 4x 8 ≤ 0
x 2 x 2 4x 2 ≤ 0
x 2x 2 4 ≤ 0
Critical numbers:
Test intervals: , 2 ⇒ x3 2x2 4x 8 < 0
Solution interval: , 2
0 1 2 3 4
−4 −3 −2 −1 0 1 2 3 4 5
−8 −6 −4 −2 0 2
2, 2 ⇒ x 3 2x 2 4x 8 < 0
2, ⇒ x 3 2x 2 4x 8 > 0
4
x 2, x 2
x
x
x

25.
27.
29.
31.
4x 2 4x 1 ≤ 0 26.
Critical number:
Test intervals:
Test: Is
Interval x-Value Value of 2x 1 Conclusion
2
1
2 ,
2x 1 2 ≤ 0
1
, 2
Solution set:
x 0
x 1
x 1
2
x 1
2
1
, 2 , 1
2 ,
2x 1 2 ≤ 0?
−2
1 2 1
1 2 1
−1
0
1
2
1
2
Positive
Positive
4x 3 6x 2 < 0 28.
2x 2 2x 3 < 0
Critical numbers:
Test intervals:
Test: Is
2x2 , 0, 0,
2x 3 < 0?
3
2, 3 2 ,
By testing an x-value in each test interval in the inequality,
we see that the solution set is: , 0 0, 3
2
Critical numbers:
Test intervals:
x 0, x 3
2
x 3 4x ≥ 0 30.
xx 2x 2 ≥ 0
x 0, x ±2
, 2, 2, 0, 0, 2, 2,
Test: Is xx 2x 2 ≥ 0?
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 2, 0 2,
x 1 2 x 2 3 ≥ 0 32.
Critical numbers:
Test intervals: , 2, 2, 1, 1, )
Test: Is
x 1, x 2
x 1 2 x 3 3 ≥ 0?
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 2,
x
Section 2.7 Nonlinear Inequalities 227
x 2 3x 8 > 0
The critical numbers are imaginary:
3 i23
±
2 2
So the set of real numbers is the solution set.
−3 −2 −1
4x 3 12x 2 > 0
4x 2 x 3 > 0
Critical numbers:
0
1 2 3
Test intervals: , 0 ⇒ 4x2x 3 < 0
Solution interval: 3,
2x 3 x 4 ≤ 0
x 3 2 x ≤ 0
Critical numbers:
x
x 0, x 3
0, 3 ⇒ 4x 2 x 3 < 0
3, ⇒ 4x 2 x 3 > 0
x 0, x 2
Test intervals: , 0 ⇒ x32 x < 0
0, 2 ⇒ x 3 2 x > 0
2, ⇒ x 3 2 x < 0
Solution intervals: , 0 2,
x 4 x 3 ≤ 0
Critical numbers:
x 0, x 3
Test intervals: , 0 ⇒ x4x 3 < 0
0, 3 ⇒ x 4 x 3 < 0
3, ⇒ x 4 x 3 > 0
Solution intervals: , 0 0, 3 or , 3

228 Chapter 2 Polynomial and Rational Functions
33.
34.
35. y 1
36.
37.
y x 2 2x 3 (a)
6
−5 7
−2
y 1
2 x2 2x 1 (a)
−10
12
−4
8x3 1
2
8
−12 12
−8
x (a)
y x 3 x 2 16x 16 (a)
48
−12 12
−24
−2 −1 0 1 2
14
x
y ≤ 0 when x ≤ 1 or x ≥ 3.
(b) y ≥ 3 when 0 ≤ x ≤ 2.
y ≤ 0 (b) y ≥ 7
1
2 x2 2x 1 ≤ 0
x 2 4x 2 ≤ 0
x 4 ± 42 412
21
4 ± 8
2
y ≥ 0 when 2 ≤ x ≤ 0, 2 ≤ x < .
(b) y ≤ 6 when x ≤ 4.
2 ± 2
y ≤ 0 when 2 2 ≤ x ≤ 2 2.
y ≤ 0 (b)
x 3 x 2 16x 16 ≤ 0
x 2 x 1 16x 1 ≤ 0
x 1x 2 16 ≤ 0
y ≤ 0 when < x ≤ 4, 1 ≤ x ≤ 4.
Solution interval:
−1
0
1
4
1
4 ,
1
1
2 x2 2x 1 ≥ 7
x 2 4x 12 ≥ 0
x 6x 2 ≥ 0
y ≥ 36
Critical numbers:
Test intervals:
1 x
Test: Is
By testing an x-value in each test interval in the inequality,
we see that the solution set is: , 1 0, 1
2
1 x
x 0, x ±1
, 1, 1, 0, 0, 1, 1,
> 0?
x
2
> 0
x
1
x > 0
x
38.
Critical numbers: x 0, x
Test intervals: , 0 ⇒
1
0, 4 1
1 4x
< 0
x
4
1
4 < 0
x
y ≥ 7 when x ≤ 2, x ≥ 6.
x 3 x 2 16x 16 ≥ 36
x 3 x 2 16x 20 ≥ 0
x 2x 5x 2 ≥ 0
y ≥ 36 when x 2, 5 ≤ x < .
⇒ 1 4x
x
⇒ 1 4x
x
, 0 1
4 ,
x
1 4x
x
> 0
< 0
< 0

39.
−2 −1 0 1 2 3 4 5
x
Section 2.7 Nonlinear Inequalities 229
x 6
2 < 0
x 1
40.
x 12
3 ≥ 0
x 2
x 6 2x 1
< 0
x 1
x 12 3x 2
≥ 0
x 2
4 x
< 0
x 1
6 2x
≥ 0
x 2
Critical numbers: x 1, x 4
Critical numbers: x 2, x 3
Test intervals: , 1, 1, 4, 4,
Test intervals: , 2 ⇒
4 x
Test: Is < 0?
x 1
6 2x
2, 3 ⇒
x 2
By testing an x-value in each test interval in the inequality,
we see that the solution set is: , 1 4,
3, ⇒
3 6 9
12 15 18
Solution interval: 2, 3
−2 −1 0 1 2 3
−2 −1
−
1
2
0 1 2
x
x
6 2x
x 2
6 2x
x 2
41.
3x 5
> 4
x 5
42.
3x 5
4 > 0
x 5
3x 5 4x 5
> 0
x 5
15 x
> 0
x 5
Critical numbers:
Critical numbers: x 5, x 15
Test intervals: , 5, 5, 15, 15,
Test intervals:
15 x
Test: Is > 0?
x 5
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 5, 15
5
x
Solution intervals: , 1
2 1,
1 1 x
, 1 ⇒ > 0
2 1 2x
1 x
1, ⇒ < 0
1 2x
, 1
x
1 x
⇒ < 0
2 1 2x 1
5 7x
< 4
1 2x
5 7x 41 2x
< 0
1 2x
1 x
< 0
1 2x
, x 1
2
> 0
< 0
< 0

230 Chapter 2 Polynomial and Rational Functions
43.
45.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 5, 3
7x 1
> 0?
x 52x 3 3
2 , 1 , 1, , 5, 5, 3 2
,
x 1, x 5, x 3
2
−5
4x 3 9x 3
x 34x 3
Critical numbers:
Test intervals:
30 5x
Test: Is
≤ 0?
x 34x 3
By testing an x-value in each test interval in the inequality,
we see that the solution set is:
−4
1
x 3
9
≤ 0
4x 3
30 5x
≤ 0
x 34x 3
−2
−4
− 3
4
0
−3
2
−2
3
−
3
2
4
x 5 >
4
x 5
1
> 0
2x 3
42x 3 x 5
x 52x 3
−1
0
1
x 3 ≤
, 3
4 , 3
4 , 3 , 3, 6, 6,
4 6 8
> 0
7x 7
> 0
x 52x 3
≤ 0
9
4x 3
x 3, x 3
, x 6
4
x
1
2x 3
x
2 1,
3
4 , 3 6,
44.
46.
Critical numbers:
Test intervals:
Solution intervals: 14, 2 6,
−14
−2
−15 −10 −5 0 5
1
x ≥
1x 3 1x
≥ 0
xx 3
3
≥ 0
xx 3
Critical numbers:
5
x 6 >
5x 2 3x 6
> 0
x 6x 2
2x 28
> 0
x 6x 2
2, 6 ⇒
6, ⇒
1
x 3
Test intervals: , 3 ⇒
3, 0 ⇒
0, ⇒
Solution intervals: , 3 0,
−4 −3 −2 −1 0 1
6
10
3
x 2
x 14, x 2, x 6
, 14 ⇒
14, 2 ⇒
x
x 3, x 0
x
2x 28
< 0
x 6x 2
2x 28
> 0
x 6x 2
2x 28
< 0
x 6x 2
2x 28
> 0
x 6x 2
3
> 0
xx 3
3
< 0
xx 3
3
> 0
xx 3

Section 2.7 Nonlinear Inequalities 231
47.
xx 2
≤ 0
x 3x 3
Critical numbers: x 0, x 2, x ±3
Test intervals:
, 3, 3, 2, 2, 0, 0, 3, 3,
xx 2
Test: Is
≤ 0?
x 3x 3
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 3, 2 0, 3
x2 2x
x2 ≤ 0
9
48.
x 3x 2
≥ 0
x
Critical numbers: x 3, x 0, x 2
x 3x 2
Test intervals: , 3 ⇒ < 0
x
x 3x 2
3, 0 ⇒ > 0
x
x 3x 2
0, 2 ⇒ < 0
x
x2 x 6
≥ 0
x
49.
51.
−3 −2 −1 0 1 2 3
5x 1 2xx 1 x 1x 1
< 0
x 1x 1
Critical numbers:
Test intervals:
x
Test: Is
By testing an x-value in each test interval in the inequality,
we see that the solution set is: , 1 2
3 , 1 3, 3x 2x 3
< 0?
x 1x 1
2
−
3
−1
0
1
y 3x
x 2
8
5x 5 2x2 2x x 2 1
x 1x 1
−6 12
−4
2
, 1, 1, 2 3 , 2 3 , 1 , 1, 3, 3,
3
x 2
, x 3, x ±1
3
4
5 2x
< 1 50.
x 1 x 1
5 2x
1 < 0
x 1 x 1
3x2 7x 6
x 1x 1
3x 2x 3
< 0
x 1x 1
x
< 0
< 0
(a) y ≤ 0 when 0 ≤ x < 2.
(b) y ≥ 6 when 2 < x ≤ 4.
2, ⇒
Solution intervals: 3, 0 2,
−3 −2 −1 0 1 2 3
Critical numbers:
Test intervals: , 4 ⇒
x
x 3x 2
x
> 0
Solution intervals: , 4, 2, 1, 6,
−4
−2
1
4, 2 ⇒
2, 1 ⇒
1, 6 ⇒
6, ⇒
0 2 4 6
x
3x
x 1 ≤
3xx 4 xx 1 3x 4x 1
≤ 0
x 1x 4
x2 4x 12
x 1x 4
x 6x 2
≤ 0
x 1x 4
x 4, x 2, x 1, x 6
x 6x 2
x 1x 4
x 6x 2
x 1x 4
x 6x 2
x 1x 4
x 6x 2
x 1x 4
≤ 0
x 6x 2
x 1x 4
x
3
x 4
> 0
< 0
< 0
< 0
> 0

232 Chapter 2 Polynomial and Rational Functions
52.
53.
55.
57.
y
y 2x2
x 2 4
−6 6
6
−2
(a) y ≥ 1 when x ≤ 2 or x ≥ 2.
This can also be expressed as x ≥ 2.
(b) y ≤ 2 for all real numbers x.
This can also be expressed as < x < .
Critical numbers:
Test intervals:
Test: Is
2x 2
x 1
14
−15 15
−6
4 x 2 ≥ 0 56.
2 x2 x ≥ 0
4 x 2 ≥ 0?
x ±2
, 2, 2, 2, 2,
By testing an x-value in each test interval in the inequality,
we see that the domain set is: 2, 2
x 2 7x 12 ≥ 0 58.
x 3x 4 ≥ 0
Critical numbers:
Test intervals:
x 3, x 4
Test: Is x 3x 4 ≥ 0?
By testing an x-value in each test interval in the inequality,
we see that the domain set is: , 3 4,
(a)
, 3, 3, 4, 4,
y ≤ 0 (b) y ≥ 8
2x 2
≤ 0
x 1
y ≤ 0 when 1 < x ≤ 2.
54. y
(a)
5x
x2 4
(b)
5x x2 4
x 2 4
x 4x 1
x2 ≥ 0
4
y ≥ 1 when 1 ≤ x ≤ 4.
y ≤ 0
y ≥ 1
5x
x2 ≥ 1
4
≥ 0
5x
x 2 4 ≤ 0 −6 6
y ≤ 0 when < x ≤ 0.
x 2 4 ≥ 0
x 2x 2 ≥ 0
Critical numbers:
x 2, x 2
Test intervals: , 2 ⇒ x 2x 2 > 0
2, 2 ⇒ x 2x 2 < 0
Domain: , 2 2,
144 9x 2 ≥ 0
94 x4 x ≥ 0
Critical numbers:
4
−4
2, ⇒ x 2x 2 > 0
Test intervals: , 4 ⇒ 94 x4 x < 0
Domain: 4, 4
2x 2
≥ 8
x 1
2x 2 8x 1
≥ 0
x 1
6x 12
≥ 0
x 1
6x 2
≥ 0
x 1
y ≥ 8 when 2 ≤ x < 1.
x 4, x 4
4, 4 ⇒ 94 x4 x > 0
4, ⇒ 94 x4 x < 0

59.
61.
x
x
x
≥ 0
x 5x 7
Critical numbers: x 0, x 5, x 7
Test intervals: , 5, 5, 0, 0, 7, 7,
x
Test: Is
≥ 0?
x 5x 7
By testing an x-value in each test interval in the inequality,
we see that the domain set is: 5, 0 7,
2 ≥ 0
60.
2x 35
0.4x 2 5.26 < 10.2 62.
0.4x 2 4.94 < 0
0.4x 2 12.35 < 0
Critical numbers: x ±3.51
Test intervals: , 3.51, 3.51, 3.51, 3.51,
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 3.51, 3.51
Section 2.7 Nonlinear Inequalities 233
x
x2 ≥ 0
9
x
≥ 0
x 3x 3
Critical numbers:
Test intervals: , 3 ⇒
3, 0 ⇒
0, 3 ⇒
3, ⇒
Domain: 3, 0 3,
1.3x 2 3.78 > 2.12
1.3x 2 1.66 > 0
x 3, x 0, x 3
x
< 0
x 3x 3
x
> 0
x 3x 3
x
< 0
x 3x 3
x
> 0
x 3x 3
Critical numbers: ±1.13
Test intervals: , 1.13, 1.13, 1.13, 1.13,
Solution set: 1.13, 1.13
63.
The zeros are x
Critical numbers: x 0.13,
x 25.13
Test intervals: , 0.13, 0.13, 25.13, 25.13,
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 0.13, 25.13
12.5 ± 12.52 0.5x
40.51.6
.
20.5
2 12.5x 1.6 > 0 64.
1.2x
Critical numbers: 4.42, 0.42
Test intervals: , 4.42, 4.42, 0.42, 0.42,
Solution set: 4.42, 0.42
2 1.2x
4.8x 2.2 < 0
2 4.8x 3.1 < 5.3
65.
1
> 3.4 66.
2.3x 5.2
1
3.4 > 0
2.3x 5.2
1 3.42.3x 5.2
> 0
2.3x 5.2
7.82x 18.68
> 0
2.3x 5.2
Critical numbers: x 2.39, x 2.26
Test intervals: , 2.26, 2.26, 2.39, 2.39,
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 2.26, 2.39
Critical numbers:
2
> 5.8
3.1x 3.7
2 5.83.1x 3.7
> 0
3.1x 3.7
23.46 17.98x
> 0
3.1x 3.7
x 1.19, x 1.30
Test intervals: , 1.19 ⇒
1.19, 1.30 ⇒
1.30, ⇒
Solution interval: 1.19, 1.30
23.46 17.98x
3.1x 3.7
23.46 17.98x
3.1x 3.7
23.46 17.98x
3.1x 3.7
> 0
< 0
< 0

234 Chapter 2 Polynomial and Rational Functions
67.
69.
s 16t 2 v 0 t s 0 16t 2 160t
(a)
(b)
16t 2 160t 0
16tt 10 0
t 0, t 10
It will be back on the ground in 10 seconds.
16t 2 160t > 384
16t 2 160t 384 > 0
16t 2 10t 24 > 0
t 2 10t 24 < 0
t 4t 6 < 0
L 2 50L 500 ≥ 0
4 < t < 6 seconds
By the Quadratic Formula we have:
Critical numbers: L 25 ± 55
Test: Is
Solution set:
2L 2W 100 ⇒ W 50 L 70.
LW ≥ 500
L50 L ≥ 500
L 2 50L 500 ≥ 0?
25 55 ≤ L ≤ 25 55
13.8 meters ≤ L ≤ 36.2 meters
68. s 16t2 v0t s0 16t 2 128t
(a) 16t2 128t 0
(b)
16tt 8 0
It will be back on the ground in 8 seconds.
16t 2 128t 128 < 0
Critical numbers: 4 22, 4 22
Test intervals:
, 4 22, 4 22, 4 22,
4 22,
Solution set:
16t 0 ⇒ t 0
t 8 0 ⇒ t 8
16t 2 128t < 128
L 2 220L 8000 ≥ 0
0 seconds ≤ t < 4 22 seconds
and 4 22 seconds < t ≤ 8 seconds
2L 2W 440 ⇒ W 220 L
By the Quadratic Formula we have:
Critical numbers:
Test:
Is L 2 220L 8000 ≥ 0?
Solution set:
LW ≥ 8000
L220 L ≥ 8000
71. R x75 0.0005x and C 30x 250,000
72. What is the price per unit?
P R C
When x 90,000:
75x 0.0005x 2 30x 250,000
0.0005x 2 45x 250,000
0.0005x 2 45x 250,000 ≥ 750,000
0.0005x 2 45x 1,000,000 ≥ 0
Critical numbers: x 40,000, x 50,000 (These were
obtained by using the Quadratic Formula.)
Test intervals:
By testing x-values
in each test interval in the inequality, we
see that the solution set is 40,000, 50,000 or
40,000 ≤ x ≤ 50,000. The price per unit is
p R
75 0.0005x.
x
P ≥ 750,000
0, 40,000, 40,000, 50,000, 50,000,
For x 40,000, p $55. For x 50,000, p $50.
Therefore, for 40,000 ≤ x ≤ 50,000, $50.00 ≤ p ≤ $55.00.
When x 100,000:
L 110 ± 1041
110 1041 ≤ L ≤ 110 1041
R $2,880,000 ⇒ 2,880,000
90,000
R $3,000,000 ⇒ 3,000,000
100,000
45.97 feet ≤ L ≤ 174.03 feet
$32 per unit
$30 per unit
Solution interval: $30.00 ≤ p ≤ $32.00

73.
C 0.0031t 3 0.216t 2 5.54t 19.1, 0 ≤ t ≤ 23
(a)
74. (a)
80
0 23
0
(b)
t C
C will be greater than 75% when
t 31, which corresponds to 2011.
24 70.5
26 71.6
28 72.9
30 74.6
32 76.8
34 79.6
(c) C 75 when t 30.41.
(b)
76. (a)
(c)
Maximum safe load
d 4 6 8 10 12
Load 2223.9 5593.9 10,312 16,378 23,792
25,000
20,000
15,000
10,000
5,000
L
2000 ≤ 168.5d 2 472.1
2472.1 ≤ 168.5d 2
14.67 ≤ d 2
3.83 ≤ d
4 6 8 10 12
Depth of the beam
The minimum depth is 3.83 inches.
d
Section 2.7 Nonlinear Inequalities 235
(d)
t C
C will be between 85% and 100%
when t is between 37 and 42. These
36 83.2 values correspond to the years 2017 to
2022.
37 85.4
38 87.8
39 90.5
40 93.5
41 96.8
42 100.4
43 104.4
(e) 85 ≤ C ≤ 100 when 36.82 ≤ t ≤ 41.89 or 37 ≤ t ≤ 42.
(f) The model is a third-degree polynomial and as
t →, C →.
75.
2R 1
R
2 R1 N 0.03t 2 9.6t 172 (b) and (d)
220 ⇒ t 5
So the number of master’s degrees earned by women
exceeded 220,000 in 1995.
N 0.03t 2 9.6t 172
320 ⇒ t 16.2
So the number of master’s degrees earned by women
will exceed 320,000 in 2006.
1 1
R R1 1
2
2R 1 2R RR 1
2R 1 R2 R 1
Since R ≥ 1, we have
2R 1
1 ≥ 0
2 R1 R 1
2R 1
≥ 1
2 R1 2
≥ 0.
2 R1 Since R1 > 0, the only critical number is R1 2. The
inequality is satisfied when R 1 ≥ 2 ohms.
Master's degrees earned
(in thousands)
320
280
240
200
160
N
2 6 10 14 18
Year (0 ↔ 1990)
N = 320
N = 220
t

236 Chapter 2 Polynomial and Rational Functions
77. True
79.
x 3 2x 2 11x 12 x 3x 1x 4
The test intervals are , 3, 3, 1, 1, 4, and
4, .
x 2 bx 4 0 80.
To have at least one real solution, b This
occurs when b ≤ 4 or b ≥ 4. This can be written as
, 4 4, .
2 16 ≥ 0.
78. True
The y-values are greater than zero for all values of x.
x 2 bx 4 0
To have at least one real solution,
b 2 414 ≥ 0
b 2 16 ≥ 0.
This inequality is true for all real values of b. Thus, the
interval for b such that the equation has at least one real
solution is , .
81.
To have at least one real solution, b2 3x
4310 ≥ 0.
2 bx 10 0 82. 2x
To have at least one real solution,
2 bx 5 0
b 2 120 ≥ 0
b 120b 120 ≥ 0
Critical numbers: b ±120 ±230
Test intervals:
, 230, 230, 230, 230,
Test: Is b
Solution set: , 230 230,
2 120 ≥ 0?
83. (a) If a > 0 and c ≤ 0, then b can be any real number. If
and then for b to be greater than or
equal to zero, b is restricted to b < 2ac or
b > 2ac.
2 a > 0 c > 0, 4ac
(b) The center of the interval for b in Exercises 79–82 is 0.
85. 4x 2 20x 25 2x 5 2 86.
87.
x 2x 2x 3
b 2 425 ≥ 0
b 2 40 ≥ 0.
This occurs when b ≤ 210 or b ≥ 210. Thus,
the interval for b such that the equation has at least
one real solution is , 210 210, .
84. (a)
(b)
x a, x b
− + +
− − +
+ − +
a
(c) The real zeros of the polynomial
x 3 2 16 x 3 4x 3 4
b
x 7x 1
x2x 3) 4x 3 x2 4x 3 88. 2x 4 54x 2xx3 27
89. Area lengthwidth 90.
2x 1x
2x 2 x
2xx 3x 2 3x 9
Area 1
2baseheight
1
2b3b 2
3
2 b2 b
x

Review Exercises for Chapter 2
1. (a) (b) y 2x2 y 2x2 (a)
Review Exercises for Chapter 2 237
Vertical stretch Vertical stretch and a reflection in the x-axis
(c) (d) y x 22 y x2 2
2. (a)
(c)
4
3
2
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
Vertical shift two units upward Horizontal shift two units to the left
4
3
1
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
y
y
x
x
y x 2 4 (b)
Vertical shift four units downward
y
3
2
−4 −3 −1
−1
−2
−5
y x 3 2
1 3 4
Horizontal shift three units to the right
5
4
3
2
1
−3 −2 −1
−1
−2
−3
y
1 2 3 4 5
x
x
4
3
2
1
−4 −3 −2 −1 1 2 3 4
(d)
−3
−4
y
4
1
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
y 4 x 2
y
Reflection in the x-axis and a vertical shift
four units upward
5
3
2
1
−4 −3 −1
−1
−2
−3
y 1
2 x 2 1
y
x
1 3 4
Vertical shrink (each y-value is multiplied by
and a vertical shift one unit downward
−4 −3 −2
4
3
2
1
−2
−3
−4
y
2 3 4
x
x
x
1
2,

238 Chapter 2 Polynomial and Rational Functions
3. gx x2 2x
5.
7.
Vertex:
x 2 2x 1 1
x 1 2 1
1, 1
Axis of symmetry:
x 1
0 x 2 2x xx 2
x-intercepts: 0, 0, 2, 0
Vertex:
−3 −2
x 2 8x 16 16 10
x 4 2 6
Axis of symmetry:
x 4 2 6
4, 6
x 4 ±6
x 4 ± 6
x 4
0 x 4 2 6
x-intercepts: 4 ± 6, 0
7
6
5
4
3
−1
−1
−2
y
−8 −4
1
2 3 4 5 6
2
−2
−4
−6
y
2
x
x
4. fx 6x x2 Vertex:
x 2 6x 9 9
x 3 2 9
3, 9
Axis of symmetry:
x 3
0 6x x 2 x6 x
x-intercepts: 0, 0, 6, 0
f x x2 8x 10 6. hx 3 4x x2 Vertex:
Axis of symmetry:
6
t-intercepts: 1 ± , 0 2
Vertex:
Axis of symmetry:
0 3 4x x 2
0 x 2 4x 3
x 2 4x 3
x 2 2 7
x 2 2 7
2, 7
4 ± 28
2
x 2
2 ± 7
x-intercepts: 2 ± 7, 0
10
8
6
4
2
−2
−2
x 2 4x 4 4 3
x 4 ± 42 413
21
ft 2t2 4t 1 8. f x x2 8x 12
2t 2 2t 1 1 1
2t 1 2 1 1
2t 1 2 3
1, 3
2t 1 2 3
0 2t 1 2 3
t 1 ± 3
2
t 1 ± 6
2
t 1
6
5
4
3
2
1
y
−3 −2 −1 1 2 3 4 5 6
t
x 2 8x 16 16 12
x 4 2 4
Vertex: 4, 4
Axis of symmetry:
0 x 2 8x 12
0 x 2x 6
x 4
x-intercepts: 2, 0, 6, 0
−2
8
6
4
2
−2
−2
−4
y
10
8
6
4
2
y
y
2 4 8 10
2 4 6 8 10
4 8
10
x
x
x

9.
11.
13.
hx 4x2 4x 13 10. f x x2 6x 1
Vertex:
4x 2 x 13
4x2 x 1 1
13
4 4
4x2 x 1
1 13
4
1
4x 2 2
12
1, 12 2
Axis of symmetry:
1
x 2 2
1
0 4x 2
3
2
12
No real zeros
x-intercepts: none
hx x 2 5x 4
x 2 5x 25
4
x 1
2
5
x 2 2
25 16
4 4
Vertex: 5 5
x 2
, 41
2 2
41
4
Axis of symmetry:
0 x 2 5x 4
4
By the Quadratic Formula, x
5 ± 41
x-intercepts: , 0 2
fx 1
3 x2 5x 4
Vertex:
25
4
x 5
2
4
20
15
10
5
−3 −2 −1 1 2 3
−8
1 5
x 3 2 2
41
4
1
3x2 5x 25 25
4
4 4
1 5
x 3 2 2
41
12
5, 41
2 12
Axis of symmetry: x 5
2
−6
−4
y
5 ± 41
.
2
−2 2
−2
−4
−10
y
0 x 2 5x 4
x
x
12.
−8
Vertex: 3, 8
Axis of symmetry:
0 x 2 6x 1
x 3 2 8
6 ± 32
2
x-intercepts: 3 ± 22, 0
Review Exercises for Chapter 2 239
x 2 6x 9 9 1
x 3
x 6 ± 62 411
21
Vertex:
1, 4 2
x- intercepts: None
3 ± 22
fx 4x 2 4x 5
1
4x 2 2
4
2
−2
−2
4x2 x 1 1 5
4 4 4
1
4x 2 2
1
−4
−6
−8
Axis of symmetry:
0 4x
4 ± 8i
By the Quadratic Formula, x
8
The equation has no real zeros.
2 4x 5
x 1
2
−8 −6 −4
−6
−4
By the Quadratic Formula, x
5 ± 41
x-intercepts: , 0 2
4
2
−2 2
−4
−6
y
5 ± 41
.
2
x
y
2
4 8
12
10
8
2
−2
−2
y
10
2 4 6
1
± i.
2
x
x

240 Chapter 2 Polynomial and Rational Functions
14.
f x 1
2 6x2 24x 22
3x 2 12x 11
3x 2 4x 4 4 11
3x 2 2 34 11
3x 2 2 1
Vertex: 2, 1
Axis of symmetry: x 2
0 3x 2 12x 11
x 12 ± 122 4311
23
3
x-intercepts: 2 ± , 0 3
15. Vertex:
Point:
Thus, fx 1
2x 4 2 1.
17. Vertex:
19. (a)
20.
Point:
2, 1 ⇒ 1 a2 4 2 1
2 4a
1
2 a
12 ± 12
6
14
12
10
8
6
4
2
y
–6 –4 –2 4 6 8 10
2 ± 3
3
4, 1 ⇒ fx ax 4 2 1 16. Vertex:
1, 4 ⇒ fx ax 1 2 4 18. Vertex:
2, 3 ⇒ 3 a2 1 2 4
1 a
Thus, fx x 1 2 4.
R 10p 2 800p
(a)
R20 $12,000
R25 $13,750
R30 $15,000
x
y
(b)
2x 2y 200
x y 100
Area xy
y 100 x
x100 x
100x x 2
x
2, 2 ⇒ fx ax 2 2 2
Point: 0, 3 ⇒ 3 a0 2 2 2
1 4a
1
4 a
fx 1
4x 2 2 2
fx 1
3x 2 2 3
3 4a 2
2, 3 ⇒ fx ax 2 2 3
Point: 1, 6 ⇒ 6 a1 2 2 3
6 9a 3
3 9a
1
3 a
(c) Area 100x x2 x 2 100x 2500 2500
x 50 2 2500
x 50 2 2500
The maximum area occurs at the vertex when x 50
and y 100 50 50. The dimensions with the
maximum area are x 50 meters and y 50 meters.
(b) The maximum revenue occurs at the vertex of the parabola.
b 800
$40
2a 210
R40 $16,000
The revenue is maximum when the price is $40 per unit.
The maximum revenue is $16,000.

21. C 70,000 120x 0.055x
The minimum cost occurs at the vertex of the parabola.
2 22.
0 0.107x2 26 0.107x
5.68x 74.5
2 5.68x 48.5
23.
26.
29.
31.
Vertex: b 120
1091 units
2a 20.055
Approximately 1091 units should be produced each day to
yield a minimum cost.
y x 3 , fx x 4 3 24.
−2
5
4
3
2
1
−3
−4
y
1 2 3 4 6 7
Transformation: Reflection in
the x-axis and a horizontal shift
four units to the right
f x 2x 24 y x4 ,
6
5
4
3
2
1
−3 −2 −1 1 2 3 4 5 6
−2
−3
y
is a shift to the right two
units and a vertical stretch of the
graph of y x4 f x
.
x
x
27.
f x 4x3 y x3 ,
3
2
1
y
−3 −2 −1
−1
−2
−3
1 2 3
x 23.7, 29.4
is a reflection in the x-axis
and a vertical stretch of the graph
of y x3 f x
.
y x 5 , fx x 3 5
−2
5
4
3
2
1
−5
1
3 4 5 6 7
Transformation: Horizontal shift
three units to the right
fx x 2 6x 9 30.
The degree is even and the leading coefficient is negative.
The graph falls to the left and falls to the right.
gx 3
4x 4 3x 2 2 32.
The degree is even and the leading coefficient is positive.
The graph rises to the left and rises to the right.
y
The age of the bride is
approximately 24 years
when the age of the groom
is 26 years.
Review Exercises for Chapter 2 241
x 5.68 ± 5.682 40.10774.5
20.107
x
x
25.
28.
−3
−2
Age of groom
27
26
25
24
23
22
y x 4 , fx 2 x 4
3
1
−1
−2
−3
y
y
20 21 22 23 24 25
Age of bride
1 2 3
Transformation: Reflection in
the x-axis and a vertical shift
two units upward
y x 5 ,
f x 1
2 x5 3
−6 −4 −2 2 4
8
6
4
y
is a vertical shrink and a
vertical shift three units upward
of the graph of y x5 f x
.
fx
The degree is odd and the leading coefficient is positive.
The graph falls to the left and rises to the right.
1
2x3 2x
hx x 5 7x 2 10x
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
6
x
x
x

242 Chapter 2 Polynomial and Rational Functions
33.
35.
37.
39.
20
f x 2x2 11x 21 34.
0 2x 2 11x 21 −9
2x 3x 7
Zeros: x all of
multiplicity 1 (odd multiplicity)
Turning points: 1
3
2 , 7,
f t t 3
3 3t 36.
0 t 3 3t
0 tt 2 3
Zeros: t 0, ±3 all of
multiplicity 1 (odd multiplicity)
Turning points: 2
f x 12x 10
3 20x2 38.
0 12x 3 20x 2
0 4x 2 3x 5
Zeros: of multiplicity 2
(even multiplicity)
x of multiplicity 1
(odd multiplicity)
5
x 0
3
Turning points: 2
−5
fx x
(a) The degree is odd and the leading coefficient is
negative. The graph rises to the left and falls to
the right.
3 x2 2 40.
(b) Zero: x 1
(c)
(d)
x 3 2 1 0 1 2
fx
(−1, 0)
−4 −3 −2
−5
−5
−40
34 10 0 2 2 6
4
3
2
1
−3
−4
y
1 2 3 4
x
−3
5
9
4
fx xx 3 2
0 xx 3 2
Zeros: x 0 of multiplicity 1
(odd multiplicity)
x 3 of multiplicity 2
(even multiplicity)
Turning points: 2
f x x 3 8x 2
0 x 3 8x 2
0 x 2 x 8
Zeros: x 0 of multiplicity 2
(even multiplicity)
x 8 of multiplicity 1
(odd multiplicity)
Turning points: 2
gx x 4 x 3 2x 2
0 x 4 x 3 2x 2
0 x 2 x 2 x 2
x 2 x 1x 2
Zeros: x 0 of multiplicity 2 (even multiplicity)
x 1 of multiplicity 1 (odd multiplicity)
x 2 of multiplicity 1 (odd multiplicity)
Turning points: 3
(a) The degree is odd and the leading coefficient, 2, is
positive. The graph rises to the right and falls to the left.
(b)
0 x
The zeros are 0 and 2.
(c)
2 0 2x
x 2
2 0 2x
x 2
3 4x2 gx 2x3 4x2 gx 2x3 4x2 x 3
2 1 0 1
(d)
gx
18
(−2, 0) (0, 0)
4
3
2
−4 −3 −1
−1
−2
−3
−4
y
1 2 3 4
x
−10
−6
−4
0 2 0 6
3
−5
10
−80
3
−3
6
10
5

41.
fx xx 3 x 2 5x 3
(a) The degree is even and the leading coefficient is
positive. The graph rises to the left and rises to
the right.
(b) Zeros:
(c)
(d)
43. (a)
45. (a)
47.
3
(−3, 0)
−4 −2 −1
x 0, 1, 3
x 4 3 2 1 0 1 2 3
fx
100 0 18 8 0 0 10 72
−15
−18
−21
y
(1, 0)
1 2 3 4
(0, 0)
fx 3x 3 x 2 3
x 3 2 1 0 1 2 3
fx
87
25
(b) The zero is in the interval 1, 0.
Zero: x 0.900
fx x 4 5x 1
x
1
3 5 23 75
x 3 2 1 0 1 2 3
fx
95 25 5 1 5 5 65
(b) There are two zeros, one in the interval 1, 0 and
one in the interval 1, 2
Zeros: x 0.200, x 1.772
24x 2 16x
Thus, 24x2 x 8
3x 2
8x 5
3x 2 ) 24x 2 x 8
15x 8
15x 10
2
8x 5
2
3x 2 .
42. hx 3x2 x4 48.
4x 8
3x 2 ) 4x 7
4x 7 4
3x 2 3
29
33x 2
Review Exercises for Chapter 2 243
(a) The degree is even and the leading coefficient, 1,
is
negative. The graph falls to the left and falls to the right.
(b)
(c)
(d)
44. (a)
46. (a)
gx 3x 2 x 4
0 3x 2 x 4
0 x 2 3 x 2
The zeros are 0, 3, and 3.
x 2 1 0 1 2
hx
4
3
2
(( 3, 0 (( 3, 0
−
4
−4 −3 −1
−1
−2
−3
−4
y
4
3
3
29
3
(0, 0)
2 0 2 4
1 3 4
f x 0.25x 3 3.65x 6.12
x
fx
6
25.98
5
(b) The only zero is in the interval 5, 4.
It is x 4.479.
f x 7x 4 3x 3 8x 2 2
(b) There are zeros in the intervals 2, 1 and 1, 0.
They are x 1.211 and x 0.509.
x
6.88
4
3
2
4.72 10.32 11.42
x 1 0 1 2 3 4
fx
9.52 6.12 2.72 0.82 1.92 7.52
x 3 2 1 0 1 2
fx
416 58 2 2 4 106

244 Chapter 2 Polynomial and Rational Functions
49.
51.
x 2 3x 1 ) 5x 3 13x 2 x 2
5x 3 15x 2 5x
Thus, 5x3 13x 2 x 2
x 2 3x 1
5 x 2 50.
2 x 2 6x 2
2x 2 6x 2
0
5x 2.
x 2 0x 2 ) x 4 3x 3 4x 2 6x 3
Thus, x4 3x 3 4x 2 6x 3
x 2 2
53. 2 6
6
Thus,
x 4 0x 3 2x 2
6x 4 4x 3 27x 2 18x
x 2
55. 4 2
57.
2
4
12
8
19
8
11
3x 3 2x 2 6x
3x 3 0x 2 6x
27
16
11
38
44
6
18
22
x 2 3x 2 52.
2 x 2 0x 3
4
2x 2 0x 4
1
x 2 3x 2 1
x 2 2 .
6x 3 8x 2 11x 4 8
x 2 .
24
24
Thus, 2x3 19x 2 38x 24
x 4
f x 20x 4 9x 3 14x 2 3x
(a)
(c)
1 20
Yes, x 1 is a zero of f.
0 20
20
20
9
0
9
9
20
11
14
0
14
0
14
11
3
Yes, x 0 is a zero of f.
3
0
3
0
8
8
2x 2 11x 6.
3
3
0
0
0
0
0
0
0
54.
56.
(b)
(d)
x 2 1 ) 3x 4 0x 3 0x 2 0x 0
3x 4 3x 2
3x4 x2 1 3x2 3 3
x2 1
3x 2 3
3x 2 0
3x 2 3
2x2 0x 1 ) 6x4 10x3 13x2 3x
5x 2
2 5x 8
6x 4 0x 3 3x 2
6x 4 10x 3 13x 2 5x 2
2x 2 1
5 0.1
0.1
0.3
0.5
0.8
0
4
4
0.1x 3 0.3x 2 0.5
x 5
3 3
3
20
9
11
29
33
4
3x 3 20x 2 29x 12
x 3
3
4
20
20
Yes, x is a zero of f.
3
1 20
20
4
9
15
24
9
20
29
14
18
14
29
15
3
10x 3 16x 2 5x
10x 3 0x 2 5x
0.5
20
19.5
No, x 1 is not a zero of f.
4
16x 2 0x 2
16x 2 0x 8
10
3x 2 5x 8 10
2x 2 1
0.1x 2 0.8x 4 19.5
x 5
12
12
0
3x 2 11x 4
3
3
0
3
15
12
0
0
0
0
12
12

58. f x 3x
(a) 4 3 8 20 16
12 16 16
3 4 4 0
3 8x2 20x 16
59.
Yes, x 4 is a zero of f.
(c)
2
3 3 8 20 16
2 4 16
3 6 24 0
f x x 4 10x 3 24x 2 20x 44
(a)
Yes, x is a zero of f.
2
3
3 1
1
10
3
7
24
21
45
Thus, f 3 421.
Thus, g2 0.
20
135
155
60. gt 2t
(a) 4 2 5 0 0
8 52 208
2 13 52 208
Thus, g4 3276.
(b) 2 2 5
22
2 5 22
5 5t4 8t 20
(e)
80
−8 5
−60
44
465
421
8
832
824
0
52 4
52 4
20
3296
3276
0
10 42
10 42
(b)
(b)
4 3
1 1
1
3
f 1 9
10
1
9
Review Exercises for Chapter 2 245
8
12
20
No, x 4 is not a zero of f.
(d) 1 3 8 20 16
3 11 9
3 11 9 25
8
102 8
102
No, x 1 is not a zero of f.
20
20
0
24
9
33
20
80
60
20
33
53
16
240
224
61. ; Factor:
(a)
Yes, is a factor of
(b)
The remaining factors of are and
(c) f x x
x 7x 1x 4
(d) Zeros: 7, 1, 4
3 4x2 x
f x 7 x 1.
25x 28
2 f x x x 4
4 1 4 25 28
4 32 28
1 8 7 0
x 4
f x.
8x 7 x 7x 1
3 4x2 25x 28 62.
(a)
Yes, is a factor of
(b)
The remaining factors are and
(c)
(d) Zeros: x
(e)
50
5
2x
2x 5 x 3.
f x 2x 5x 3x 6
2 , 3, 6
2 f x 2x
6 2 11 21 90
12 6 90
2 1 15 0
x 6
f x.
x 15 2x 5x 3
3 11x2 21x 90
−7 5
−100
44
53
9

246 Chapter 2 Polynomial and Rational Functions
63. f x x
Factors: x 2, x 3
(a) 2 1
4 4x3 7x2 22x 24 64. f x x
(a) 2 1 11 41 61 30
1
2
9
18
23
46
15
30
0
4 11x3 41x2 61x 30
(b)
(c)
(d) Zeros:
(e)
Both are factors since the remainders are zero.
x
The remaining factors are x 1 and x 4.
2 3 1 6 5 12
3 9 12
1 3 4 0
3x 4 x 1x 4
f x x 1x 4x 2x 3
−3
1
2, 1, 3, 4
40
−10
4
2
6
7
12
5
5
22
10
12
24
24
0
(b)
Yes, and are both factors of
x
The remaining factors are x 1 and x 3.
2 5 1 9 23 15
5 20 15
1 4 3 0
x 2 x 5
f x.
4x 3 x 1x 3
(c)
(d) Zeros:
(e)
f x x 1x 3x 2x 5
x 1, 2, 3, 5
4
−6 12
65. 6 4 6 2i 66. 3 25 3 5i 67. i 2 3i 1 3i
68. 5i i 2 1 5i 69. 7 5i 4 2i 7 4 5i 2i 3 7i
70. 2 2
2 2 2 2 2 2 2 2 2
i i i
2 2 2 2 2 2 2 2 i
−8
i 2i
71. 5i13 8i 65i 40i2 40 65i 72. 1 6i5 2i 5 2i 30i 12i2 73.
75.
4 46i
5 28i 12
17 28i
10 8i2 3i 20 30i 16i 24i2 74. i6 i3 2i i18 12i 3i 2i2 6 i 6 i 4 i
4 i 4 i 4 i
24 10i i2
16 1
23 10i
17
23 10
17 17 i
76.
17 7i
26
17 7i
26 26
i20 9i
20i 9i 2
9 20i
3 2i 3 2i 5 i
5 i 5 i 5 i
15 3i 10i 2i2
25 i 2

77.
79.
4 2 4 2 3i 2 1 i
2 3i 1 i 2 3i 2 3i 1 i 1 i
3x 2 1
x 2 1
3
8 12i
4 9
8
13
x ± 1
3
12
i 1 i
13
8
13 1 12
i i 13
21 1
13 13 i
± 1
i ±3
3 3 i
2 2i
1 1
3x 2 1 0 80.
78.
Review Exercises for Chapter 2 247
1 5 1 4i 52 i
2 i 1 4i 2 i1 4i
2 8x 2 0
8x 2 2
x 2 1
4
x ± 1
2 i
81. x2 2x 10 0 82. 6x2 3x 27 0
83.
87.
89.
x 2 2x 1 10 1
x 1 2 9
x 1 ±9
x 1 ± 3i
f x 3xx 2 2 84.
Zeros: x 0, x 2
f x x 4x 6x 2ix 2i
Zeros: x 4, x 6, x 2i, x 2i
Possible rational zeros:
f x x 4x 9 2 85.
Zeros: x 9, 4
± 1
±1, ±3, ±5, ±15, ±
3 5 15
4 , ± 4 , ± 4 , ± 4
1 3 5 15
2 , ± 2 , ± 2 , ± 2 ,
88.
x b ± b2 4ac
2a
3 ± 639
12
3 ± 3i71
12
1 4i 10 5i
2 8i i 4i 2
9 i 2 9i
2 9i 2 9i
18 81i 2i 9i2
4 81i 2
9 83i
85
3 ± 32 4627
26
x 1x 8
Zeros: x 1, x 8
9 83i
85 85
1 71
±
4 4 i
f x x 2 9x 8 86.
Zeros: x 5, 8, 3 ± i
f x x 3 6x
xx 2 6
Zeros: x 0, ±6i
f x x 8x 5 2 x 3 ix 3 i
fx 4x3 8x2 3x 15 90. fx 3x4 4x3 5x2 8
Possible rational zeros: ± 1 2 4 8
±1, ±2, ±4, ±8, 3 , ± 3 , ± 3 , ± 3

248 Chapter 2 Polynomial and Rational Functions
91.
93.
95.
96.
f x x 3 2x 2 21x 18 92.
Possible rational zeros: ±1, ±2, ±3, ±6, ±9, ±18
1 1
1
2
1
3
21
3
18
18
18
0
x 3 2x 2 21x 18 x 1x 2 3x 18
x 1x 6x 3
The zeros of f x are x 1, x 6, and x 3.
f x x 3 10x 2 17x 8 94.
Possible rational zeros: ±1, ±2, ±4, ±8
1 1
1
10
1
9
17
9
8
8
8
0
x 3 10x 2 17x 8 x 1x 2 9x 8
x 1x 1x 8
x 1 2 x 8
The zeros of f x are x 1 and x 8.
f x x 4 x 3 11x 2 x 12
Possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12
3 1
1
4 1
1
1
3
4
4
4
0
11
12
1
1
0
1
4
4
0
x 4 x 3 11x 2 x 12 x 3x 4x 2 1
1
3
4
12
12
0
The real zeros of f x are x 3, and x 4.
f x 25x 4 25x 3 154x 2 4x 24
3 25
25
2 25
25
75
50
50
50
So, f x 25x 4 25x3 154x2 25 0 4 0
4x 24
x 3x 225x 2 4
x 3x 25x 25x 2.
Zeros: x 3, 2, ± 2
5
154
150
4
4
0
8
8
4
12
8
24
24
0
f x 3x 3 20x 2 7x 30
Possible rational zeros:
1 3
3
20
3
23
So, f x 3x 3 20x 2 7x 30
x 13x 2 23x 30
x 13x 5x 6
0 x 13x 5x 6.
Zeros: x 1, 5
3 , 6
7
23
30
±1, ±2, ±3, ±5, ±6, ±10, ±15,
±30, ± 1 2 5
3 , ± 3 , ± 3 ,
30
30
0
f x x 3 9x 2 24x 20
± 10
3
Possible rational zeros: ±1, ±2, ±4, ±5, ±10, ±20
5 1
9
5
So, f x x3 9x2 1 4 4 0
24x 20
x 5x 2 4x 4
x 5x 2 2 .
Zeros: x 5, 2
Possible rational zeros:
± 6
25, ± 8
25, ± 12
25, ± 24
± 25
4
± 25,
1
25, ± 2
25, ± 3
± 25,
24
±24, ±
5 ,
1
5, ± 2
5, ± 3
5, ± 4
5, ± 6
5, ± 8
5, ± 12
±1, ±2, ±3, ±4, ±6, ±8, ±12,
5 ,
24
20
20
20

97. Since is a zero, so is
3x 2x 4x Multiply by 3 to clear the fraction.
2 fx 3x 3i
3i.
3
2
3x 4x 3ix 3i
3x 2 14x 8x 2 3
3x 4 14x 3 17x 2 42x 24
Review Exercises for Chapter 2 249
Note: where a is any real nonzero number, has zeros 2
, 4, and ±3i.
f x a3x4 14x3 17x2 42x 24,
98. Since is a zero and the coefficients are real,
must also be a zero.
x2 x 6x 12 1 2i 99. fx x Zero: i
1 2i
Since i is a zero, so is i.
f x x 2x 3x 1 2ix 1 2i
i 1 4 1 4
4
i 1 4i 4
3 4x2 x 4,
x 2 x 6x 2 2x 5
x 4 x 3 3x 2 17x 30
4i 1
4i 1
1
2 4i
4i
1 2 0
hx x 4ix 4ix 2
Zeros: x ± 4i, 2
2
4i
2 4i
16
16 8i
8i
8i
8i
32
32
0
1
i 1
1
4 i
4 i
i
4
4i
4i
4i
fx x ix ix 4, Zeros: x ± i, 4
100. hx x
Since 4i is a zero, so is 4i.
3 2x2 16x 32 101. gx 2x Zero: 2 i
Since 2 i is a zero, so is 2 i
4 3x3 13x2 37x 15,
102.
2 i 2
2
2 i 2
3
4 2i
1 2i
1 2i
4 2i
gx x 2 ix 2 i2x2 2 5 3 0
5x 3
Zeros: x 2 ± i, 1
2 , 3
fx 4x4 11x3 14x2 6x 103. fx x3 4x2 5x
x4x 3 11x 2 14x 6
One zero is x 0. Since 1 i is a zero, so is 1 i.
1 i 4
4
1 i 4
4
11
4 4i
7 4i
7 4i
4 4i
3
f x xx 1 ix 1 i4x 3
xx 1 ix 1 i4x 3
Zeros: 0, 3
4 , 1 i, 1 i
14
11 3i
3 3i
3 3i
3 3i
0
6
6
0
0
x 2 ix 2 i2x 1x 3
xx 2 4x 5
xx 5x 1
Zeros: x 0, 5, 1
3
13
5i
13 5i
0
13 5i
10 5i
37
31 3i
6 3i
6 3i
6 3i
15
15
0

250 Chapter 2 Polynomial and Rational Functions
104.
gx x 3 7x 2 36
2 1
1
7
2
9
0
18
18
36
36
0
The zeros of x are x 3, 6. The zeros of gx are 2, 3, 6.
gx x 2x 3x 6
2 9x 18 x 3x 6
105. gx x Zero: x 4
4 4x3 3x2 40x 208, 106.
4 1
1
4 1
1
4
4
0
0
4
4
3
0
3
3
16
13
40
12
52
52
52
gx x 4 2 x 2 4x 13
By the Quadratic Formula the zeros of x are
x 2 ± 3i. The zeros of gx are x 4 of multiplicity
2, and x 2 ± 3i.
2 4x 13
gx x 4 2 x 2 3ix 2 3i
x 4 2 x 2 3ix 2 3i
0
208
208
107. gx 5x
gx has two variations in sign, so g has either two or no
positive real zeros.
3 3x2 6x 9
gx 5x 3 3x 2 6x 9
gx has one variation in sign, so g has one negative
real zero.
4
3
1
4
4
1
5
3
5
4
17
4
Since the last row entries alternate in sign,
x is a lower bound.
1
4
0
fx x 4 8x 3 8x 2 72x 153
3 1
1
3 1
1
8
3
11
11
3
8
8
33
41
41
24
72
123
17
51
51
51
By the Quadratic Formula, the zeros of x are
2 8x 17
x 8 ± 82 4117
21
153
153
The zeros of fx are 3, 3, 4 i, 4 i.
fx x 3x 3x 4 ix 4 i
0
8 ± 4
2
4 ± i.
108. hx 2x
hx has three variations in sign, so h has either three or
one positive real zeros.
5 4x3 2x2 5
hx 2x 5 4x 3 2x 2 5
2x 5 4x 3 2x 2 5
hx has two variations in sign, so h has either two or
no negative real zeros.
109.
(a)
Since the last row has all positive entries,
is an upper bound.
(b) 1
fx 4x
1 4 3 4 3
4 1 5
4 1 5 2
x 1
4 4
3 3x2 4x 3 110. gx 2x
(a) 8 2 5 14 8
16 88 592
2 11 74 600
Since the last row has all positive entries, x 8 is an
upper bound.
(b) 4 2 5
8
14
52
8
152
3 5x2 14x 8
2 13 38 144
Since the last row entries alternate in sign, x 4 is a
lower bound.

111.
114.
117.
f x
Domain: all real numbers
except x 12
5x
x 12
f x x2 x 2
x 2 4
Domain: all real numbers
hx
2x 10
x 2 2x 15
2x 5
x 3x 5
2
, x 5
x 3
Vertical asymptote:
x 3
Horizontal asymptote: y 0
x
112.
119. fx
(a) Domain: all real numbers x except x 0
5
x2 (b) No intercepts
(c) Vertical asymptote: x 0
Horizontal asymptote: y 0
(d)
x
±3
5
9
1
−1 1 2
−2
−3
y
±2
y 5
5
4
±1
x
f x 3x2
1 3x
1 3x 0
3x 1
x 1
3
Domain: all real numbers
except x 1
3
115. fx
Vertical asymptote: x 3
4
x 3
Horizontal asymptote: y 0
x
Review Exercises for Chapter 2 251
113.
f x
8
x 2 10x 24
8
x 4x 6
Domain: all real numbers
except x 4 and x 6
116. fx
Vertical asymptote: none
2x2 5x 3
x2 2
Horizontal asymptote: y 2
118. hx
Vertical asymptotes: x 2, x 1
x3 4x2 x2 3x 2 x2x 4
x 2x 1
Horizontal asymptotes: none
120. fx
(a) Domain: all real numbers x except x 0
4
x
(b) No intercepts
(c) Vertical asymptote: x 0
Horizontal asymptote: y 0
(d)
x
y
3
4
3
4
3
2
1
−3 −2 −1 1 2 3 4
−2
−3
y
2
2
1
4
x
1 2 3
4 2
4
3
x

252 Chapter 2 Polynomial and Rational Functions
2 x 2
121. gx x
1 x x 1
(a) Domain: all real numbers x except x 1
(b) x-intercept:
y-intercept:
(c) Vertical asymptote: x 1
Horizontal asymptote: y 1
(d)
6
4
(0, 2)
(−2, 0)
−2
−4
−6
−8
y
2, 0
0, 2
x 0 2 3
y 2 5
1
1
4
2
123. px
(a) Domain: all real numbers x
x2
x2 1
(b) Intercept:
(c) Horizontal asymptote:
(d)
4
3
2
−3 −2 −1 (0, 0) 2 3
−2
y
2
0, 0
x
2
y 1
x ±3 ±2 ±1 0
9
10
4
5
1
2
y 0
125. fx
(a) Domain: all real numbers x
(d)
x
x2 1
(b) Intercept:
0, 0
(c) Horizontal asymptote: y 0
x
122.
x 3
hx
x 2
(a) Domain: all real numbers x except x 2
(b) x-intercept: 3, 0
3
y-intercept: 0, 2
(c) Vertical asymptote: x 2
Horizontal asymptote: y 1
(d)
x 1 0 1 3 4 5
y 2 0
(( 0, 3
2
5
4
3
−2 −1
−2
1 4
(3, 0)
5 6
−3
4
3
y
3
2
124. f x
(a) Domain: all real numbers x
2x
x2 4
(b) Intercept:
(c) Horizontal asymptote:
(d)
x 2
1 0 1 2
2
5
1
2
y 0
3
2
1
(0, 0)
−1
−2
−3
0, 0
y
1 2 3
x
x
1
2
y 0
x 2 1 0 1 2
y 0
1
2
1
2
2
5
2
5
2
1
(0, 0)
−1
−2
2
5
y
1
2
2
3
1 2
x

4
126. hx
x 1
(a) Domain: all real numbers x except x 1
2
(b) y- intercept: 0, 4
(c) Vertical asymptote: x 1
Horizontal asymptote: y 0
(d)
x 2 1 0 2 3 4
4
9
y 1 4 4 1
7
6
5
(0, 4)
3
1
y
−3 −2 −1 2 3 4 5
128. y
(a) Domain: all real numbers x except x ±2
2x2
x2 4
(b) Intercept:
(c) Vertical asymptotes: x 2, x 2
Horizontal asymptote: y 2
(d)
(0, 0)
6
4
0, 0
x ±5 ±4 ±3 ±1 0
50
21
y 0
−6 −4 4 6
y
8
3
18
5
x
x
2
3
4
9
127. f x
(a) Domain: all real numbers x
6x2
x2 1
129.
(b) Intercept:
(c) Horizontal asymptote:
(d)
−6
−4
−2
Review Exercises for Chapter 2 253
4
2
−8
0, 0
y
(0, 0)
2 4
6
y 6
x ±3 ±2 ±1 0
27
5
y 3 0
24
5
f x 6x2 11x 3
3x 2 x
(c) Vertical asymptote: x 0
Horizontal asymptote: y 2
(d)
(a) Domain: all real numbers except and
(b) intercept:
y-intercept:
none
3
x
x-
, 0 2 1
x x 0
3
x
y
3x 12x 3
x3x 1
2
7
2
2
−8 −6 −4 −2
−2
−4
−6
−8
4
3
, 0
2
6 8
y
1
((
x
2x 3
, x
x
1
3
1 2 3 4
5 1 1
x
1
2
5
4

254 Chapter 2 Polynomial and Rational Functions
130.
132.
f x 6x2 7x 2
4x 2 1
2x 13x 2
2x 12x 1
(a) Domain: all real numbers x except
(b) y- intercept: 0, 2
x-intercept:
(c) Vertical asymptote:
Horizontal asymptote: y
(d)
3
x
2
1
2
x 3 2 1 0
2
1 2
y
11
5
2
−3 −2 −1
2 3
2
, 0
3
f x x2 1
x 1
y
2
, 0 3
8
3
((
3x 2 1
, x
2x 1 2
5 2 0
x
3
x ± 1
2
(a) Domain: all real numbers x except x 1
(b) y-intercept:
(c) Vertical asymptote:
(d)
Using long division,
Slant asymptote:
−6 −4 −2 2 4 6
y
0, 1
4
(0, 1)
x 1
1
3
4
5
f x x2 1
2
x 1
x 1 x 1 .
y x 1
x 6
2
0 4
y 5
1
37
5
3
2
13
2
x
1
2
5
2
17
5
131. fx
(a) Domain: all real numbers x
2x3
x2 2x
2x
1 x2 1
(b) Intercept:
(c) Slant asymptote:
(d)
−3
−2
3
2
1
(0, 0)
−1
−2
−3
0, 0
y
1 2
y 2x
x 2 1 0 1 2
y 1 0 1
16
5
3
x
16
5

133.
135.
fx 3x3 2x 2 3x 2
3x 2 x 4
x 1 23
, x 1
3 3x 4
(a) Domain: all real numbers x except
(b) x- intercepts: 1, 0 and
y-intercept:
(c) Vertical asymptote:
Slant asymptote: y x
(d)
1
x
3
4
3
x 3 2 0 1 2 3
y
((
C C
x
0, − 1
2
3x 2x 1x 1
3x 4x 1
3x 2x 1
3x 4
44
13
4
3
2
1
y
2
, 0
3
(1, 0)
−2 −1
2 3 4
−2
((
0, 1 2
12
5
1
2
2
, 0 3
x
0 2
x 1, x 4
3
14
5
134.
0.5x 500
, 0 < x 136.
x
Horizontal asymptote: C
As x increases, the average cost per unit approaches the
horizontal asymptote, C 0.5 $0.50.
0.5
0.5
1
Review Exercises for Chapter 2 255
fx 3x3 4x 2 12x 16
3x 2 5x 2
(a) Domain: all real x except x 2 or
(b) intercept:
intercepts: 4
, 0 , 2, 0
3 x-
y- 0, 8
(c) Vertical asymptote:
(d)
(a)
x 2x 23x 4
x 23x 1
x 23x 4
, x 2
3x 1
Using long division,
f x 3x2 10x 8
3x 1
Slant asymptote:
x
y
4
96
13
4
2
y
((
4
, 0
3
−6 −4 −2
4 6
−2 (2, 0)
−6
(0, −8)
1
21
4
x 1
3
y x 3
x 3
0 1 2 4
8
C 528p
, 0 ≤ p < 100
100 p
4000
0
0
x
1
2
0
x 1
3
5
3x 1 .
16
11
(b) When C million.
52825
p 25,
$176
100 25
100
When p 50, C million.
52850
$528
100 50
When p 75, C million.
(c) As p → 100, C → . No, it is not possible.
52875
$1584
100 75

256 Chapter 2 Polynomial and Rational Functions
137. (a)
y
2 in. 2 in.
(b) The area of print is x 4y 4, which is
30 square inches.
0
0
x 4y 4 30
x
2 in.
2 in.
y 4 30
x 4
y 30
4
x 4
y
y
y
30 4x 4
x 4
4x 14
x 4
22x 7
x 4
22x 7 2x2x 7
Total area xy x x 4
x 4
100
(c) Because the horizontal margins total 4 inches, x must be
greater than 4 inches. The domain is x > 4.
(d)
200
4 32
0
The minimum area occurs when x 9.477 inches, so
y
22 9.477 7
9.477 4
9.477 inches.
The least amount of paper used is for a page size of about
9.48 inches by 9.48 inches.
138.
The limiting amount of uptake is determined
by the horizontal asymptote,
y
90
18.47
0.23 80.3 mgdm2 18.47x 2.96
y , 0 < x
0.23x 1
139.
CO2 hr.
Critical numbers:
Test intervals: ,
Test: Is 3x 42x 1 < 0?
4
x
3 , , 43
4
6x
3x 42x 1 < 0
1
3 , x 2
2 6x
5x 4 < 0
2 5x < 4
140.
2 x 2 x ≥ 15 141.
2 x 2 x 15 ≥ 0
Critical numbers: x
Test intervals: , 3 ⇒ 2x 5x 3 > 0
5
2x 5x 3 ≥ 0
2 , x 3
5 2 , 3,
⇒ 2x 5x 3 > 0
5
2 ⇒ 2x 5x 3 < 0
Solution interval: , 3 5
2 ,
1
, 2, 1 2 ,
By testing an x-value in each test interval in the
inequality, we see that the solution set is: 4
3
x 3 16x ≥ 0
xx 4x 4 ≥ 0
Critical numbers:
Test intervals:
x 0, x ±4
Test: Is xx 4x 4 ≥ 0?
By testing an x- value in each test interval in the inequality,
we see that the solution set is: 4, 0 4, .
, 1
2
, 4, 4, 0, 0, 4, 4,

142.
144.
146.
148.
12x 3 20x 2 < 0 143.
4x 2 3x 5 < 0
Critical numbers:
Test intervals:
x 0, x 5
3
, 0 ⇒ 12x 3 20x 2 < 0
Solution interval: , 0 0, 5
3
5
3 , ⇒ 12x3 20x2 0,
> 0
5
3 ⇒ 12x3 20x2 < 0
x 5
< 0 145.
3 x
Critical numbers:
x 5, x 3
Test intervals: , 3 ⇒
3, 5 ⇒
5, ⇒
x 5
< 0
3 x
x 5
> 0
3 x
x 5
< 0
3 x
Solution intervals: , 3 5,
1 1
>
x 2 x
1 1
> 0
x 2 x
Critical numbers:
Test intervals:
x 2, x 0
, 0 ⇒ 1 1
> 0
x 2 x
0, 2 ⇒ 1 1
< 0
x 2 x
2, ⇒ 1 1
> 0
x 2 x
Solution interval: , 0 2,
P
2000 ≤
10001 3t
5 t
10001 3t
5 t
20005 t ≤ 10001 3t
10,000 2000t ≤ 1000 3000t
1000t ≤ 9000
t ≥ 9 days
147.
Review Exercises for Chapter 2 257
2x 1 3x 1
≤ 0
x 1x 1
2x 2 3x 3
≤ 0
x 1x 1)
x 5
≤ 0
x 1x 1
Critical numbers: x 5, x ±1
Test intervals: , 5, 5, 1, 1, 1, 1,
x 5
Test: Is
≤ 0?
x 1x 1
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 5, 1 1,
2
x 1 ≤
3
x 1
x 4x 3
≥ 0
x
Critical numbers: x 4, x 3, x 0
Test intervals: , 4, 4, 3, 3, 0, 0,
x 4x 3
Test: Is
≥ 0?
x
By testing an x-value in each test interval in the inequality,
we see that the solution set is: 4, 3 0,
x2 7x 12
≥ 0
x
50001 r 2 > 5500
149. False. A fourth-degree polynomial
can have at most four zeros and
complex zeros occur in conjugate
pairs.
1 r 2 > 1.1
1 r > 1.0488
r > 0.0488
r > 4.9%
150. False. (See Exercise 123.)
The domain of
f x 1
x 2 1
is the set of all real numbers x.

258 Chapter 2 Polynomial and Rational Functions
151. The maximum (or minimum) value of a quadratic
function is located at its graph’s vertex. To find the
vertex, either write the equation in standard form or
use the formula
1.
fx ax 3 bx 2 cx d
x k) ax3 bx2 ax
cx d
2 ak bx ak2 bk c
ax 3 akx 2
ak bx 2 cx
ak bx 2 ak 2 bkx
ak 2 bk cx d
ak 2 bk cx ak 3 bk 2 ck
ak 3 bk 2 ck d
Thus, and
Since the remainder r ak f k r.
3 bk2 f k ak ck d,
3 bk2 f x ax
ck d.
3 bx2 cx d x kax2 ak bx ak2 bx c ak3 bk2 ck d
2. (a)
b b
, f
2a 2a .
If the leading coefficient is positive, the vertex is a
minimum. If the leading coefficient is negative, the
vertex is a maximum.
Problem Solving for Chapter 2
(b)
(c)
y
y 3 y 2
1 2
2 12
3 36
4 80
5 150
6 252
7 392
8 576
9 810
10 1100
x
2 3
x
2 2
36 ⇒ x
1
8
3 ⇒ x 6
2 x3 1
8 2x2 1
8 288
x3 2x2 288; a 1, b 2 ⇒ a2
x
1
b3 8
3 x2 252 ⇒ x 6
152. Answers will vary. Sample answer:
Polynomials of degree n > 0
with real coefficients can
be written as the product of linear and quadratic factors
with real coefficients, where the quadratic factors have
no real zeros.
153. An asymptote of a graph is a line to which the graph becomes arbitrarily close as x increases or decreases without bound.
(d)
(e)
(f)
(g)
Setting the factors equal to zero and solving for the
variable can find the zeros of a polynomial function.
To solve an equation is to find all the values of the
variable for which the equation is true.
3x3 x2 90; a 3, b 1 ⇒ a2
b3 9
93x 3 9x 2 990
3x 3 3x 2 810 ⇒ 3x 9 ⇒ x 3
2x3 5x2 2500; a 2, b 5 ⇒ a2 4
b3
125
4
125 2x3 4
125 5x2 4
125 2500
2x
5 3
2x
5 2
80 ⇒ 2x
4 ⇒ x 10
5
7x3 6x2 1728; a 7, b 6 ⇒ a2 49
b3 216
49
216 7x3 49
216 6x2 49
216 1728
7x
6 3
7x
6 2
392 ⇒ 7x
7 ⇒ x 6
6
10x3 3x2 297; a 10, b 3 ⇒ a2 100
b3 27
100
27 10x3 100
27 3x2 100
27 297
10x
3 3
10x
3 2
1100 ⇒ 10x
10 ⇒ x 3
3

3.
V l w h x 2 x 3
x 2 x 3 20
x 3 3x 2 20 0
Possible rational zeros: ±1, ±2, ±4, ±5, ±10, ±20
2 1
x 2
1
3
2
5
or x
0
10
10
x 2x 2 5x 10 0
f x rx
4. False. Since we have qx
dx dx .
f x dxqx rx,
f x f 1
The statement should be corrected to read since qx
x 1 x 1 .
f 1 2
4 1 3a
3 3a
5 ± 15i
2
a 1
b 1 41 5
y x 2 5x 4
20
20
0
(b) Enter the data points
and use the regression feature to obtain
y x2 0, 4, 1, 0, 2, 2, 4, 0,
6, 10
5x 4.
Problem Solving for Chapter 2 259
5. (a)
1, 0: 0 a1
4 a b
4 a 1 4a
2 4, 0: 0 a4
0 16a 4b 4 44a b 1
0 4a b 1 or b 1 4a
b1 4
2 0, 4: 4 a0
4 c
b4 4
2 y ax
b0 c
2 bx c 6. (a)
9 4
Slope 5
3 2
Slope of tangent line is less than 5.
(b)
4 1
Slope 3
2 1
Slope of tangent line is greater than 3.
4.41 4
(c) Slope 4.1
2.1 2
Slope of tangent line is less than 4.1.
f 2 h f 2
(d) Slope
2 h 2
x
x
x + 3
Choosing the real positive value for x we have: x 2
and x 3 5.
The dimensions of the mold are 2 inches 2 inches 5 inches.
(e)
2 h2 4
h
4h h2
h
4 h, h 0
Slope 4 h, h 0
4 1 3
4 1 5
4 0.1 4.1
The results are the same as in (a)–(c).
(f) Letting h get closer and closer to 0, the slope
approaches 4. Hence, the slope at 2, 4 is 4.

260 Chapter 2 Polynomial and Rational Functions
7.
9.
10.
f x x kqx r 8. (a)
(a) Cubic, passes through 2, 5, rises to the right
a bia bi a 2 abi abi b 2 i 2 a 2 b 2
Since a and b are real numbers, a is also a real number.
2 b2 fx
Vertical asymptote:
Horizontal asymptote:
(i)
(ii)
(iii)
(iv)
One possibility:
f x x 2x 2 5
x 3 2x 2 5
(b) Cubic, passes through 3, 1, falls to the right
One possibility:
f x x 3x 2 1
x 3 3x 2 1
ax b
cx d
x d
c
y a
c
a > 0, b < 0, c > 0, d < 0
Both the vertical asymptote and the horizontal
asymptote are positive. Matches graph (d).
a > 0, b > 0, c < 0, d < 0
Both the vertical asymptote and the horizontal
asymptote are negative. Matches graph (b).
a < 0, b > 0, c > 0, d < 0
The vertical asymptote is positive and the horizontal
asymptote is negative. Matches graph (a).
a > 0, b < 0, c > 0, d > 0
The vertical asymptote is negative and the horizontal
asymptote is positive. Matches graph (c).
(b)
(c)
z m 1
z
1 1 1 i
1 i 1 i 1 i
z m 1
z
z m 1
z
1 i
2
1 1 3 i
3 i 3 i 3 i
3 i
10
1
2 8i
1 2 8i
2 8i 2 8i
2 8i
68
1 1
2 2 i
3 1
10 10 i
1
2
34 17 i
11.
(a) is a vertical asymptote.
a causes a vertical stretch if and a vertical
shrink if For a > 1, the graph
becomes wider as a increases. When a is negative
the graph is reflected about the x-axis.
0 < fx
b 0 ⇒ x b
a > 1
a < 1.
ax
x b2 (b) a 0. Varying the value of b varies the vertical
asymptote of the graph of f. For b > 0, the graph
is translated to the right. For b < 0, the graph is
reflected in the x-axis and is translated to the left.

Problem Solving for Chapter 2 261
12. (a)
(b)
50
1
y
0.007x 0.44
0
0
70
1
y 0.0313x
0.007x 0.44
y 2 Age, x Near point, y
50 (c)
Age, x
Near point, y Quadratic Rational
16 3.0
Model Model
32
44
4.7
9.8
0
0
70
16
32
3.0
4.7
3.66
2.32
3.05
4.63
50 19.7
44 9.8 11.83 7.58
60 39.4
50 19.7 19.97 11.11
60 39.4 38.54 50.00
1.586x 21.02
The models are fairly good fits to the data. The quadratic
model seems to be a better fit for older ages and the
rational model a better fit for younger ages.
(d) For x 25, the quadratic model yields y 0.9325 inches
and the rational model yields y 3.774 inches.
(e) The reciprocal model cannot be used to predict the near
point for a person who is 70 years old because it results in
a negative value y 20. The quadratic model yields
y 63.37 inches.

262 Chapter 2 Polynomial and Rational Functions
Chapter 2 Practice Test
1. Sketch the graph of f x x and identify the vertex and the intercepts.
2 6x 5
2. Find the number of units x that produce a minimum cost C if
C 0.01x 2 90x 15,000.
3. Find the quadratic function that has a maximum at 1, 7 and passes through the point 2, 5.
4. Find two quadratic functions that have x-intercepts and 4
2, 0 , 0.
5. Use the leading coefficient test to determine the right and left end behavior of the graph
of the polynomial function fx 3x 5 2x 3 17.
6. Find all the real zeros of f x x 5 5x 3 4x.
7. Find a polynomial function with 0, 3, and 2 as zeros.
8. Sketch f x x 3 12x.
9. Divide 3x using long division.
4 7x 2 2x 10 by x 3
10. Divide x 3 11 by x 2 2x 1.
11. Use synthetic division to divide 3x 5 13x 4 12x 1 by x 5.
12. Use synthetic division to find f6 given fx 7x 3 40x 2 12x 15.
13. Find the real zeros of fx x 3 19x 30.
14. Find the real zeros of fx x 4 x 3 8x 2 9x 9.
15. List all possible rational zeros of the function fx 6x 3 5x 2 4x 15.
16. Find the rational zeros of the polynomial fx x 3 20
3 x2 9x 10
3 .
17. Write fx x as a product of linear factors.
4 x3 5x 10
18. Find a polynomial with real coefficients that has 2, 3 i,
and 3 2i as zeros.
3

19. Use synthetic division to show that 3i is a zero of fx x 3 4x 2 9x 36.
x 1
20. Sketch the graph of fx and label all intercepts and asymptotes.
2x
21. Find all the asymptotes of fx 8x2 9
x 2 1 .
22. Find all the asymptotes of fx 4x2 2x 7
.
x 1
23. Given z1 4 3i and z 2 2 i, find the following:
(a) z1 z2 (b)
z 1 z 2
(c) z 1z 2
24. Solve the inequality: x 2 49 ≤ 0
25. Solve the inequality:
x 3
≥ 0
x 7
Practice Test for Chapter 2 263