Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Cumulative Review Exercises - Page 952: 15


The partial decomposition is, $\frac{8}{x-3}+\frac{-2}{x-2}+\frac{-3}{x+2}$

Work Step by Step

Consider the given partial decomposition $\frac{3{{x}^{2}}+17x-38}{\left( x-3 \right)\left( x-2 \right)\left( x+2 \right)}=\frac{A}{x-3}+\frac{B}{x-2}+\frac{C}{x+2}$ Multiplying both sides by $\left( x-3 \right)\left( x-2 \right)\left( x+2 \right)$ we get, $\begin{align} & 3{{x}^{2}}+17x-38=A\left( x-2 \right)\left( x+2 \right)+B\left( x-3 \right)\left( x+2 \right)+C\left( x-3 \right)\left( x-2 \right) \\ & 3{{x}^{2}}+17x-38=A\left( {{x}^{2}}-4 \right)+B\left( {{x}^{2}}-x-6 \right)+C\left( {{x}^{2}}-5x+6 \right) \\ & 3{{x}^{2}}+17x-38=\left( A+B+C \right){{x}^{2}}+\left( -B-5C \right)x+\left( -4A-6B+6C \right) \\ \end{align}$ Therefore, equating the equal powers of $x$, we get, $\begin{align} & A+B+C=3 \\ & -B-5C=17 \\ & -4A-6B+6C=-38 \end{align}$ Adding $A+B+C=3$ and $-B-5C=17$ $\begin{align} & A-4C=20 \\ & A=4C+20 \end{align}$ From $-B-5C=17$ $B=-17-5C$ Therefore, form $A+B+C=3$ to get, $\begin{align} & -4\left( 4C+20 \right)-6\left( -17-5C \right)+6C=-38 \\ & -16C-80+102+30C+6C=-38 \\ & 20C=-60 \\ & C=-3 \end{align}$ Therefore $\begin{align} & A=-4\times 3+20=8 \\ & B=-17+5\times 3=-2 \end{align}$ Therefore, the required partial fraction is $\frac{3{{x}^{2}}+17x-38}{\left( x-3 \right)\left( x-2 \right)\left( x+2 \right)}=\frac{8}{x-3}+\frac{-2}{x-2}+\frac{-3}{x+2}$
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