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: 20


The partial fraction is, $\frac{3}{x}+\frac{-5}{\left( x-2 \right)}+\frac{4}{\left( x+2 \right)}$

Work Step by Step

$\begin{align} & \frac{2{{x}^{2}}-18x-12}{{{x}^{3}}-4x}=\frac{2{{x}^{2}}-18x-12}{x\left( {{x}^{2}}-{{2}^{2}} \right)} \\ & =\frac{2{{x}^{2}}-18x-12}{x\left( x-2 \right)\left( x+2 \right)} \\ & =\frac{A}{x}+\frac{B}{\left( x-2 \right)}+\frac{C}{\left( x+2 \right)} \end{align}$ Now, multiply $x\left( x-2 \right)\left( x+2 \right)$ on both sides: $\begin{align} & x\left( x-2 \right)\left( x+2 \right)\frac{2{{x}^{2}}-18x-12}{x\left( x-2 \right)\left( x+2 \right)}=x\left( x-2 \right)\left( x+2 \right)\frac{A}{x}+x\left( x-2 \right)\left( x+2 \right)\frac{B}{\left( x-2 \right)}+x\left( x-2 \right)\left( x+2 \right)\frac{C}{\left( x+2 \right)} \\ & 2{{x}^{2}}-18x-12=\left( x-2 \right)\left( x+2 \right)A+x\left( x+2 \right)B+x\left( x-2 \right)C \\ & =\left( {{x}^{2}}+2x-2x-4 \right)A+B{{x}^{2}}+2Bx+C{{x}^{2}}-2Cx \\ & =A{{x}^{2}}-4A+B{{x}^{2}}+2Bx+C{{x}^{2}}-2Cx \end{align}$ $2{{x}^{2}}-18x-12={{x}^{2}}\left( A+B+C \right)+x\left( 2B-2C \right)-4A$ …… (1) Then, compare the coefficient of equation (1): $A+B+C=2$ …… (2) $2B-2C=-18$ $B-C=-9$ …… (3) $-4A=-12$ …… (4) $A=3$ …… (5) And put the value of A in equation (2): $3+B+C=2$ $B+C=-1$ …… (6) And add equation (6) with equation (3): $B=-5$ Then, put the value of B in equation (6): $\begin{align} & -5+C=-1 \\ & C=-1+5 \\ & C=4 \end{align}$ Therefore, $\frac{2{{x}^{2}}-18x-12}{{{x}^{3}}-4x}=\frac{3}{x}+\frac{-5}{\left( x-2 \right)}+\frac{4}{\left( x+2 \right)}$ Thus, the partial fraction of the given expression is $\frac{3}{x}+\frac{-5}{\left( x-2 \right)}+\frac{4}{\left( x+2 \right)}$.
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