## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 842: 54

#### Answer

In order to find the partial function decomposition of the rational fraction with distinct linear factors in the denominator, the expression is equated to the expression $\frac{A}{ax+b}+\frac{B}{cx+d}$ and then the values of A and B are calculated by comparing the terms of the equations.

#### Work Step by Step

In order to find the partial function decomposition of the rational fraction with distinct linear factors in the denominator, the expression is equated to the expression $\frac{A}{ax+b}+\frac{B}{cx+d}$ and then the values of A and B are calculated by comparing the equations. Let us take an example: Consider $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}$ And the partial fraction decomposition of the rational fraction is as follows: $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}=\frac{A}{x+2}+\frac{B}{x+4}$ And multiply both sides by $\left( x+2 \right)\left( x+4 \right)$: \begin{align} & 2x=A\left( x+4 \right)+B\left( x+2 \right) \\ & 2x=Ax+4A+Bx+2B \\ & 2x=\left( A+B \right)x+4A+2B \end{align} And equate the equation: $A+B=2$ (I) $4A+2B=0$ (II) Solving the above equation, From (I), $A=2-B$, put in equation (II) : \begin{align} & 4\left( 2-B \right)+2B=0 \\ & 8-4B+2B=0 \\ & 8-2B=0 \\ & 2B=8 \end{align} $B=4$ Putting the value of $B=4$ in equation (I) \begin{align} & A+4=2 \\ & A=2-4 \\ & A=-2 \end{align} This implies, $A=-2,B=4$ For these values of $A$ and $B$ in $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}=\frac{A}{x+2}+\frac{B}{x+4}$. Thus, $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}=-\frac{2}{x+2}+\frac{4}{x+4}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.