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

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