College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.3 - Page 550: 49

Answer

$\dfrac{ax+b}{(x-c)^2}=\dfrac{a}{x-c}+\dfrac{ac+b}{(x-c)^2}$

Work Step by Step

We are given the fraction: $\dfrac{ax+b}{(x-c)^2}$ As the denominator is already factored we can write the partial fraction decomposition: $\dfrac{ax+b}{(x-c)^2}=\dfrac{A}{x-c}+\dfrac{B}{(x-c)^2}$ Multiply all terms by the least common denominator $(x-c)^2$: $(x-c)^2\cdot\dfrac{ax+b}{(x-c)^2}=(x-c)^2\cdot\dfrac{A}{x-c}+(x-c)^2\cdot\dfrac{B}{(x-c)^2}$ $ax+b=A(x-c)+B$ $ax+b=Ax-Ac+B$ $ax+b=Ax+(-Ac+B)$ Identify the coefficients: $\begin{cases} A=a\\ -Ac+B=b \end{cases}$ $A=a$ $-ac+B=b$ $B=ac+b$ The partial fraction decomposition is: $\dfrac{ax+b}{(x-c)^2}=\dfrac{a}{x-c}+\dfrac{ac+b}{(x-c)^2}$
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