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 841: 15


The partial fraction is $\frac{4}{7\left( x-3 \right)}-\frac{8}{7\left( 2x+1 \right)}$.

Work Step by Step

We find the partial fraction of the stated expression as given below, $\begin{align} & \frac{4}{2{{x}^{2}}-5x-3}=\frac{4}{\left( x-3 \right)\left( 2x+1 \right)} \\ & =\frac{A}{\left( x-3 \right)}+\frac{B}{\left( 2x+1 \right)} \end{align}$ Now, multiply $\left( x-3 \right)\left( 2x+1 \right)$ on both sides: $\begin{align} & \left( x-3 \right)\left( 2x+1 \right)\frac{4}{\left( x-3 \right)\left( 2x+1 \right)}=\left( x-3 \right)\left( 2x+1 \right)\frac{A}{\left( x-3 \right)}+\left( x-3 \right)\left( 2x+1 \right)\frac{B}{\left( 2x+1 \right)} \\ & 4=\left( 2x+1 \right)A+\left( x-3 \right)B \\ & =2Ax+A+Bx-3B \\ & 4=x\left( 2A+B \right)+A-3B \end{align}$ …… (I) Then, compare the coefficient of equation (I): $2A+B=0$ …… (II) $A-3B=4$ …… (III) Then, multiply equation (II) by 3 and add with equation (III), Then, $\begin{align} & 7A=4 \\ & A=\frac{4}{7} \end{align}$ Now, put the value of A in equation (II): $\begin{align} & 2\left( \frac{4}{7} \right)+B=0 \\ & B=-\frac{8}{7} \end{align}$ Therefore, $\frac{4}{2{{x}^{2}}-5x-3}=\frac{4}{7\left( x-3 \right)}-\frac{8}{7\left( 2x+1 \right)}$ Thus, the partial fraction of the provided expression is $\frac{4}{7\left( x-3 \right)}-\frac{8}{7\left( 2x+1 \right)}$.
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