College Algebra, 2013a

8.6 Partial Fraction Decomposition 633
gives us x = ± √ 2, and each of these zeros must be of multiplicity one since Theorem 3.14
enables us to now factor x 2 − 2= ( x − √ 2 )( x + √ 2 ) . Hence,
8x
x 2 − 2 =
8x
(
x −
√
2
)(
x +
√
2
) =
A
x − √ 2 +
B
x + √ 2
Clearing fractions, we get 8x = A ( x + √ 2 ) + B ( x − √ 2 ) or 8x =(A + B)x +(A − B) √ 2.
We get the system
{
A + B = 8
(A − B) √ 2 = 0
From (A − B) √ 2 = 0, we get A = B, which, when substituted into A + B = 8 gives B =4.
Hence, A = B =4andweget
4x 3
x 2 − 2 =4x + 8x
x 2 − 2 =4x + 4
x + √ 2 + 4
x − √ 2
5. At first glance, the denominator D(x) =x 4 +6x 2 + 9 appears irreducible. However, D(x) has
three terms, and the exponent on the first term is exactly twice that of the second. Rewriting
D(x) = ( x 2) 2 +6x 2 + 9, we see it is a quadratic in disguise and factor D(x) = ( x 2 +3 ) 2 .
Since x 2 + 3 clearly has no real zeros, it is irreducible and the form of the decomposition is
x 3 +5x − 1
x 4 +6x 2 +9 = x3 +5x − 1
(x 2 +3) 2 = Ax + B
x 2 +3 + Cx + D
(x 2 +3) 2
When we clear denominators, we find x 3 +5x − 1=(Ax + B) ( x 2 +3 ) + Cx+ D which yields
x 3 +5x − 1=Ax 3 + Bx 2 +(3A + C)x +3B + D. Our system is
⎧
⎪⎨
⎪⎩
A = 1
B = 0
3A + C = 5
3B + D = −1
We have A = 1 and B =0fromwhichwegetC = 2 and D = −1. Our final answer is
x 3 +5x − 1
x 4 +6x 2 +9 =
x
x 2 +3 + 2x − 1
(x 2 +3) 2
6. Once again, the difficulty in our last example is factoring the denominator. In an attempt to
get a quadratic in disguise, we write
x 4 +16= ( x 2) 2
+4 2 = ( x 2) 2
+8x 2 +4 2 − 8x 2 = ( x 2 +4 ) 2
− 8x
2