## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 9 - 9.4 - Partial Fractions - 9.4 Exercises - Page 668: 22

#### Answer

$[\frac{1}{6}][\frac{-1}{2x+3} + \frac{1}{2x-3}]$

#### Work Step by Step

We must first find the Partial Fraction Decomposition: $\frac{A}{2x+3} + \frac{B}{2x-3} = \frac{1}{(2x+3)(2x-3)}$ We must then solve for the constants: $A(2x-3) + B(2x+3) = 1$ $2Ax - 3A + 2Bx + 3B = 1$ $2Ax + 2Bx = 0$ $-3A + 3B = 1$ This can be represented with the following matrix: $\begin{bmatrix} 2 & 2 & |0\\ -3 & 3 & |1\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & |0\\ -3 & 3 & |1\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & |0\\ 0 & 6 & |1\\ \end{bmatrix}$ $6B = 1$ $B = \frac{1}{6}$ $A + \frac{1}{6} = 0$ $A = \frac{-1}{6}$ The Partial Fraction is: $[\frac{1}{6}][\frac{-1}{2x+3} + \frac{1}{2x-3}]$ Checking the Result: $[\frac{1}{6}]\frac{-2x+3+2x+3}{(2x+3)(2x-3)} = [\frac{1}{6}]\frac{6}{(2x+3)(2x-3)} = \frac{1}{(2x+3)(2x-3)}$ Therefore, the answer is correct.

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