Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 842: 49


The partial fraction is $\frac{a}{x-c}+\frac{ac+b}{{{\left( x-c \right)}^{2}}}$.

Work Step by Step

We know that the partial fraction can be written as given below, $\begin{align} & \frac{ax+b}{{{\left( x-c \right)}^{2}}}=\frac{A}{x-c}+\frac{B}{{{\left( x-c \right)}^{2}}} \\ & =\frac{A\left( x-c \right)+B}{{{\left( x-c \right)}^{2}}} \end{align}$ And, $ax+b=A\left( x-c \right)+B$ …… (1) Substitute the value of $x=c$ in equation (1), $ac+b=B$ Again, put the value of $x=0$ in equation (1), $\begin{align} & b=-Ac+B \\ & b=-Ac+ac+b \\ & Ac=ac \\ & A=a \end{align}$ Therefore, $\frac{ax+b}{{{\left( x-c \right)}^{2}}}=\frac{a}{x-c}+\frac{ac+b}{{{\left( x-c \right)}^{2}}}$ Thus, the partial fraction decomposition form is $\frac{ax+b}{{{\left( x-c \right)}^{2}}}=\frac{a}{x-c}+\frac{ac+b}{{{\left( x-c \right)}^{2}}}$.
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