## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Review Exercises - Page 877: 17

#### Answer

The partial fraction decomposition of the rational expression is $\frac{6}{x-4}+\frac{5}{x+3}$

#### Work Step by Step

Factor the denominator by the method of grouping as given below: \begin{align} & {{x}^{2}}-x-12={{x}^{2}}-4x+3x-12 \\ & =x\left( x-4 \right)+3\left( x-4 \right) \\ & =\left( x-4 \right)\left( x-3 \right) \end{align} Set up the partial fraction decomposition by writing an unknown constant as given below: $\frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\frac{A}{x-4}+\frac{B}{x+3}$ Multiply both sides of the equation by $\left( x-4 \right)\left( x+3 \right)$ as given below: \begin{align} & \left( x-4 \right)\left( x+3 \right)\cdot \frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\left( x-4 \right)\left( x+3 \right)\left( \frac{A}{x-4}+\frac{B}{x+3} \right) \\ & \left( x-4 \right)\left( x+3 \right)\cdot \frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\left( x-4 \right)\left( x+3 \right)\left( \frac{A}{x-4} \right)+\left( x-4 \right)\left( x+3 \right)\left( \frac{B}{x+3} \right) \end{align} And divide the common factors: \begin{align} & \left( x+3 \right)\left( x-4 \right)\cdot \frac{11x-2}{\left( x+3 \right)\left( x-4 \right)}=\left( x-4 \right)\left( x+3 \right)\left( \frac{A}{x-4} \right)+\left( x-4 \right)\left( x+3 \right)\left( \frac{B}{x+3} \right) \\ & 11x-2=\left( x+3 \right)A+\left( x-4 \right)B \\ & 11x-2=Ax+3A+Bx-4B \\ & 11x-2=\left( A+B \right)x+\left( 3A-4B \right) \end{align} Then, equate the coefficients of like terms of the equation to write a system of equations as given below: $A+B=11$ (I) $3A-4B=-2$ (II) Multiply equation (I) by 4 and add it with equation (II) to get the value of A: \begin{align} & +\text{ }\underline{\begin{align} & 4A+4B=44 \\ & 3A-4B=-2 \\ \end{align}} \\ & \text{ 7}A=42 \\ \end{align} Divide both sides by 7: \begin{align} & \frac{7A}{7}=\frac{42}{7} \\ & A=6 \end{align} Hence, the value of $A$ is $6$. Then, substitute the value of $A$ in equation (I) so that the value of $B$ can be computed: \begin{align} & A+B=11 \\ & 6+B=11 \\ & B=11-6 \\ & B=5 \end{align} Put the values of $A$ and $B$ in the given equation and determine the partial fraction decomposition: \begin{align} & \frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\frac{A}{x-4}+\frac{B}{x+3} \\ & =\frac{6}{x-4}+\frac{5}{x+3} \end{align} Hence, the partial fraction decomposition of the rational expression is $\frac{6}{x-4}+\frac{5}{x+3}$.

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