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: 34


The partial fraction is, $\frac{14}{3\left( x-1 \right)}+\frac{4}{{{\left( x-1 \right)}^{2}}}+\frac{\left( -14x+4 \right)}{3\left( {{x}^{2}}+2 \right)}$.

Work Step by Step

$\frac{10{{x}^{2}}+2x}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}=\frac{A}{\left( x-1 \right)}+\frac{B}{{{\left( x-1 \right)}^{2}}}+\frac{Cx+D}{\left( {{x}^{2}}+2 \right)}$. Now, take the L.C.M. on the right side: $\frac{10{{x}^{2}}+2x}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}=\frac{A\left( x-1 \right)\left( {{x}^{2}}+2 \right)+B\left( {{x}^{2}}+2 \right)+\left( Cx+D \right){{\left( x-1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}$. By eliminating the denominator from both sides, we get $\begin{align} & 10{{x}^{2}}+2x=A\left( x-1 \right)\left( {{x}^{2}}+2 \right)+B\left( {{x}^{2}}+2 \right)+\left( Cx+D \right){{\left( x-1 \right)}^{2}} \\ & =A\left( {{x}^{3}}+2x-{{x}^{2}}-2 \right)+B{{x}^{2}}+2B+\left( Cx+D \right)\left( {{x}^{2}}+1-2x \right) \\ & =A{{x}^{3}}-A{{x}^{2}}+2Ax-2A+B{{x}^{2}}+2B+C{{x}^{3}}+Cx-2C{{x}^{2}}+D{{x}^{2}}+D-2Dx \\ & =\left( A+C \right){{x}^{3}}+\left( -A+B-2C+D \right){{x}^{2}}+\left( 2A+C-2D \right)x-2A+2B+D \end{align}$ Then, compare the coefficients of ${{x}^{3}},\ {{x}^{2}},{{x}^{1}}$ and the constant term: $A+C=0$ …… (I) $-A+B-2C+D=10$ …… (II) $2A+C-2D=2$ …… (III) $-2A+2B+D=0$ …… (IV) Then, find the value of C from equation (I): $C=-A$ …… (V) And the value of C is put in equation (III): $2A-A-2D=2$ $A-2D=2$ …… (VI) Now, find the value of B from equation (II): $B=10+A+2C-D$ …… (VII) Substitute the value of C in equation (VII): $B=10+A+2\left( -A \right)-D$ $B=10-A-D$ …… (VIII) Also, put the value of B in equation (IV): $\begin{align} & -2A+2\left( 10-A-D \right)+D=0 \\ & -2A+20-2A-2D+D=0 \end{align}$ $4A+D=20$ …… (IX) Also, multiply equation (IX) by 2 and add with equation (VI): $\begin{align} & 8A+2D=40 \\ & A-2D=2 \\ & 9A=42 \\ & A=\frac{42}{9}=\frac{14}{3} \end{align}$ And put the value of A in equation (I): $\begin{align} & \frac{14}{3}+C=0 \\ & C=\frac{-14}{3} \end{align}$ Also, put the value of A in equation (VI): $\begin{align} & \frac{14}{3}-2D=2 \\ & 2D=\frac{14}{3.}-2 \\ & 2D=\frac{14-6}{3} \\ & D=\frac{4}{3} \end{align}$ Also, substitute the values of A and D in equation (VIII): $\begin{align} & B=10-\frac{14}{3}-\frac{4}{3} \\ & =\frac{30-14-4}{3} \\ & =\frac{12}{3} \\ & =4 \end{align}$ Then, $\frac{10{{x}^{2}}+2x}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}=\frac{\frac{14}{3}}{\left( x-1 \right)}+\frac{4}{{{\left( x-1 \right)}^{2}}}+\frac{\frac{-14}{3}x+\frac{4}{3}}{\left( {{x}^{2}}+2 \right)}$. Thus, the partial fraction of the provided expression is $\frac{14}{3\left( x-1 \right)}+\frac{4}{{{\left( x-1 \right)}^{2}}}+\frac{\left( -14x+4 \right)}{3\left( {{x}^{2}}+2 \right)}$.
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