## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 14

#### Answer

The partial fraction is, $\frac{6}{\left( x-3 \right)}+\frac{3}{\left( x+5 \right)}$

#### Work Step by Step

We find the partial fraction of the decomposition as given below, \begin{align} & \frac{9x+21}{{{x}^{2}}+2x-15}=\frac{9x+21}{\left( x-3 \right)\left( x+5 \right)} \\ & =\frac{A}{\left( x-3 \right)}+\frac{B}{\left( x+5 \right)} \end{align} Now, multiply $\left( x-3 \right)\left( x+5 \right)$ on both sides: \begin{align} & \left( x-3 \right)\left( x+5 \right)\frac{9x+21}{\left( x-3 \right)\left( x+5 \right)}=\left( x-3 \right)\left( x+5 \right)\frac{A}{\left( x-3 \right)}+\left( x-3 \right)\left( x+5 \right)\frac{B}{\left( x+5 \right)} \\ & 9x+21=\left( x+5 \right)A+\left( x-3 \right)B \\ & =Ax+5A+Bx-3B \\ & 9x+21=x\left( A+B \right)+5A-3B \end{align} …… (1) Then, compare the coefficient of equation (1): $A+B=9$ …… (2) $5A-3B=21$ …… (3) Then, multiply equation (2) by 3 and add with equation (3), Then, \begin{align} & A=\frac{48}{8} \\ & =6 \end{align} Now, put the value of A in equation (II): \begin{align} & 6+B=9 \\ & B=9-6 \\ & =3 \end{align} Therefore, $\frac{9x+21}{{{x}^{2}}+2x-15}=\frac{6}{\left( x-3 \right)}+\frac{3}{\left( x+5 \right)}$ Thus, the partial fraction of the given expression is $\frac{6}{\left( x-3 \right)}+\frac{3}{\left( x+5 \right)}$.

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