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: 27


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

Work Step by Step

Now, take L.C.M of both sides: $\frac{{{x}^{2}}}{{{\left( x-1 \right)}^{2}}\left( x+1 \right)}=\frac{A\left( x-1 \right)\left( x+1 \right)+B\left( x+1 \right)+C{{\left( x-1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}\left( x+1 \right)}$ By eliminating the denominators from both sides, we get: $\begin{align} & {{x}^{2}}=A\left( x-1 \right)\left( x+1 \right)+B\left( x+1 \right)+C{{\left( x-1 \right)}^{2}} \\ & =A\left( {{x}^{2}}-1 \right)+Bx+B+C\left( {{x}^{2}}+1-2x \right) \\ & =A{{x}^{2}}-A+Bx+B+C{{x}^{2}}+C-2Cx \\ & =\left( A+C \right){{x}^{2}}+\left( B-2C \right)x-A+B+C \end{align}$ Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+C=1$ …… (I) $B-2C=0$ …… (II) $-A+B+C=0$ …… (III) Then, find the value of A from equation (I): $A=1-C$ …… (IV) Also, find the value of B from equation (II): $B=2C$ ……(V) And put the equation (V) and equation (IV) in equation (III): $\begin{align} & -\left( 1-C \right)+2C+C=0 \\ & -1+C+3C=0 \\ & 4C=1 \\ & C=\frac{1}{4} \end{align}$ Then, put the value of C in equation (V) and equation (IV): $\begin{align} & A=1-\left( \frac{1}{4} \right) \\ & =1-\frac{1}{4} \\ & =\frac{3}{4} \end{align}$ $\begin{align} & B=2\left( \frac{1}{4} \right) \\ & =\frac{1}{2} \end{align}$ And put the value of A, B and C in the given expression: $\begin{align} & \frac{{{x}^{2}}}{{{\left( x-1 \right)}^{2}}\left( x+1 \right)}=\frac{A}{\left( x-1 \right)}+\frac{B}{{{\left( x-1 \right)}^{2}}}+\frac{C}{\left( x+1 \right)} \\ & \frac{{{x}^{2}}}{{{\left( x-1 \right)}^{2}}\left( x+1 \right)}=\frac{\frac{3}{4}}{\left( x-1 \right)}+\frac{\frac{1}{2}}{{{\left( x-1 \right)}^{2}}}+\frac{\frac{1}{4}}{\left( x+1 \right)} \end{align}$ Therefore,, $\frac{{{x}^{2}}}{{{\left( x-1 \right)}^{2}}\left( x+1 \right)}=\frac{\frac{3}{4}}{\left( x-1 \right)}+\frac{\frac{1}{2}}{{{\left( x-1 \right)}^{2}}}+\frac{\frac{1}{4}}{\left( x+1 \right)}$ Thus, the partial fraction of the provided expression is $\frac{3}{4\left( x-1 \right)}+\frac{1}{2{{\left( x-1 \right)}^{2}}}+\frac{1}{4\left( x+1 \right)}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.