Precalculus (6th Edition) Blitzer

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

Chapter 7 - Review Exercises - Page 877: 21


The partial fraction decomposition of the rational expression is $\frac{6}{5\left( x-2 \right)}-\frac{\left( 6x+3 \right)}{5\left( {{x}^{2}}+1 \right)}$

Work Step by Step

By putting a constant $ A $ over the linear factor $ x-2$ and a linear expression $ Bx+C $ over the quadratic factor: $\frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{\left( x-2 \right)}+\frac{Bx+C}{{{x}^{2}}+1}$ And multiply both sides of the equation by $\left( x-2 \right)\left( {{x}^{2}}+1 \right)$ as given below: $\begin{align} & \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\left( x-2 \right)\left( {{x}^{2}}+1 \right)\left( \frac{A}{\left( x-2 \right)}+\frac{Bx+C}{{{x}^{2}}+1} \right) \\ & \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\left\{ \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{A}{x-2}+\left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{Bx+C}{{{x}^{2}}+1} \right\} \end{align}$ And divide out the common factors: $\begin{align} & \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\left\{ \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{A}{\left( x-2 \right)}+\left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{Bx+C}{{{x}^{2}}+1} \right\} \\ & 3x=\left\{ \left( {{x}^{2}}+1 \right)\cdot A+\left( x-2 \right)\cdot \left( Bx+C \right) \right\} \\ & 3x=A{{x}^{2}}+A+\left( B{{x}^{2}}-2Bx+Cx-2C \right) \\ & 3x=\left( A+B \right){{x}^{2}}+\left( -2B+C \right)x+\left( A-2C \right) \end{align}$ Then, equate the coefficients of like terms of the equation to write a system of equations as shown below: $ A+B=0$ (I) $-2B+C=3$ (II) $ A-2C=0$ (III) From equations (I) and (III), it is obtained that $ A=-B\text{ and }A=2C $. Put $ A=-B $ in equation (III): $\begin{align} & A-2C=0 \\ & -B-2C=0 \end{align}$ (IV) Multiply by 2 in equation (II) and add with equation (IV): $\begin{align} & +\text{ }\underline{\begin{align} & -4B+2C=6 \\ & -B-2C=0 \end{align}} \\ & \text{ }-5B=6 \\ \end{align}$ Divide both sides by $-5$ as given below: $\begin{align} & \left( \frac{-5B}{-5} \right)=\frac{6}{-5} \\ & B=-\frac{6}{5} \end{align}$ So, the value of $ B $ is $-\frac{6}{5}$. Then, put the value of $ B $ in equation (IV) to get the value of $ C $: $\begin{align} & -B-2C=0 \\ & -\left( -\frac{6}{5} \right)-2C=0 \\ & 2C=-\frac{6}{5} \\ & C=-\frac{6}{10} \end{align}$ As $ A=-B $, $ A=\frac{6}{5}$ Put the values of $ A,\,\,B,\text{ and }C $ in the given equation, and define the partial fraction decomposition: $\begin{align} & \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{\left( x-2 \right)}+\frac{Bx+C}{{{x}^{2}}+1} \\ & =\frac{\frac{6}{5}}{\left( x-2 \right)}+\frac{-\frac{6}{5}x-\frac{6}{10}}{{{x}^{2}}+1} \\ & =\frac{6}{5\left( x-2 \right)}+\frac{\frac{-12x-6}{10}}{{{x}^{2}}+1} \\ & =\frac{6}{5\left( x-2 \right)}-\frac{\not{2}\left( 6x+3 \right)}{1\not{0}\left( {{x}^{2}}+1 \right)} \\ & =\frac{6}{5\left( x-2 \right)}-\frac{\left( 6x+3 \right)}{5\left( {{x}^{2}}+1 \right)} \end{align}$ Thus, the partial fraction decomposition of the rational expression is $\frac{6}{5\left( x-2 \right)}-\frac{\left( 6x+3 \right)}{5\left( {{x}^{2}}+1 \right)}$.
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