## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 9 - 9.4 - Partial Fractions - 9.4 Exercises - Page 668: 21

#### Answer

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

#### Work Step by Step

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

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