College Algebra 9th txtbk

A-22 APPENDIX B SPECIAL TOPICS
Therefore,
5x 2 8x 5
(x 2)(x 2 x 1) 3
x 2 2x 1
x 2 x 1
7x 2 11x 6
MATCHED PROBLEM 3 Decompose into partial fractions: .
(x 1)(2x 2 3x 2)
EXAMPLE 4 Repeating Quadratic Factors
x 3 4x 2 9x 5
Decompose into partial fractions: .
(x 2 2x 3) 2
SOLUTION
Because x 2 2x 3 can’t be factored further in the real numbers, we proceed to use part
4 from Theorem 3 to write
So for all x,
x 3 4x 2 9x 5
Ax B
(x 2 2x 3) 2 x 2 2x 3 Cx D
(x 2 2x 3) 2
x 3 4x 2 9x 5 (Ax B)(x 2 2x 3) Cx D
Because the substitution of carefully chosen values of x doesn’t lead to the immediate determination
of A, B, C, or D, we multiply and rearrange the right side to obtain
x 3 4x 2 9x 5 Ax 3 (B 2A)x 2 (3A 2B C)x (3B D)
Now we use Theorem 1 to equate coefficients of terms of like degree:
A 1
B 2A 4
3A 2B C 9
3B D 5
(Ax B)(x2 2x 3) Cx D
(x 2 2x 3) 2
1x 3
4x 2 9x 5
Ax 3 (B 2A)x 2 (3A 2B C )x (3B D)
From these equations we easily find that A 1, B 2, C 2, and D 1. Now we can
write
x 3 4x 2 9x 5 x 2
(x 2 2x 3) 2 x 2 2x 3 2x 1
(x 2 2x 3) 2
3x 3 6x 2 7x 2
MATCHED PROBLEM 4 Decompose into partial fractions: .
(x 2 2x 2) 2
ANSWERS TO MATCHED PROBLEMS
4
1. 2.
x 2 3
x 3
2
3. 4.
x 1 3x 2
2x 2 3x 2
3
x 1 2
x 2 1
(x 2) 2
3x
x 2 2x 2 x 2
(x 2 2x 2) 2