## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 12

#### Answer

The partial fraction is, $\frac{3}{\left( x-2 \right)}+\frac{2}{\left( x+1 \right)}$

#### Work Step by Step

We have to apply the partial fraction decomposition as given below, $\frac{5x-1}{\left( x-2 \right)\left( x+1 \right)}=\frac{A}{\left( x-2 \right)}+\frac{B}{\left( x+1 \right)}$ Now, multiply $\left( x-2 \right)\left( x+1 \right)$ on both sides, \begin{align} & \left( x-2 \right)\left( x+1 \right)\frac{5x-1}{\left( x-2 \right)\left( x+1 \right)}=\left( x-2 \right)\left( x+1 \right)\frac{A}{\left( x-2 \right)}+\left( x-2 \right)\left( x+1 \right)\frac{B}{\left( x+1 \right)} \\ & 5x-1=\left( x+1 \right)A+\left( x-2 \right)B \\ & =Ax+A+Bx-2B \\ & 5x-1=x\left( A+B \right)+A-2B \end{align} …… (I) Then, compare the coefficient of equation (I), $A+B=5$ …… (II) $A-2B=-1$ …… (III) Then, subtract equation (II) from equation (III), Then, $-3B=-6$ \begin{align} & B=\frac{-6}{-3} \\ & =2 \end{align} Now, put the value in equation (II), \begin{align} & A+2=5 \\ & A=5-2 \\ & =3 \end{align} Therefore, $\frac{5x-1}{\left( x-2 \right)\left( x+1 \right)}=\frac{3}{\left( x-2 \right)}+\frac{2}{\left( x+1 \right)}$ Thus, the partial fraction of the given expression is $\frac{3}{\left( x-2 \right)}+\frac{2}{\left( x+1 \right)}$.

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