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: 13


The partial fraction is $\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$.

Work Step by Step

Let us consider the decomposition as, $\begin{align} & \frac{7x-4}{{{x}^{2}}-x-12}=\frac{7x-4}{\left( x+3 \right)\left( x-4 \right)} \\ & =\frac{A}{\left( x+3 \right)}+\frac{B}{\left( x-4 \right)} \end{align}$ Then, multiply $\left( x+3 \right)\left( x-4 \right)$ on both sides: $\begin{align} & \left( x+3 \right)\left( x-4 \right)\frac{7x-4}{\left( x+3 \right)\left( x-4 \right)}=\left( x+3 \right)\left( x-4 \right)\frac{A}{\left( x+3 \right)}+\left( x+3 \right)\left( x-4 \right)\frac{B}{\left( x-4 \right)} \\ & 7x-4=\left( x-4 \right)A+\left( x+3 \right)B \\ & =Ax-4A+Bx+3B \\ & 7x-4=x\left( A+B \right)-4A+3B \end{align}$ …… (I) Now, compare the coefficient of equation (I): $A+B=7$ …… (II) $-4A+3B=-4$ …… (III) Then, multiply equation (II) by 4 and add with equation (III): $\begin{align} & 7B=24 \\ & B=\frac{24}{7} \end{align}$ Now, put the value of B in equation (II): $\begin{align} & A+\frac{24}{7}=7 \\ & A=7-\frac{24}{7} \\ & =\frac{49-24}{7} \\ & =\frac{25}{7} \end{align}$ Therefore, $\frac{7x-4}{{{x}^{2}}-x-12}=\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$ Thus, the partial fraction of the given expression is $\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$.
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