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: 25


The partial fraction is, $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$

Work Step by Step

$\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{{{\left( x-1 \right)}^{2}}}$ Now, multiply both sides by $\begin{align} & {{x}^{2}}+2x+7=A{{\left( x-1 \right)}^{2}}+Bx\left( x-1 \right)+Cx \\ & =A\left( {{x}^{2}}+1-2x \right)+B{{x}^{2}}-Bx+Cx \\ & =A{{x}^{2}}+A-2Ax+B{{x}^{2}}-Bx+Cx \\ & =\left( A+B \right){{x}^{2}}+\left( -2A-B+C \right)x+A \end{align}$ Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+B=1$ …… (1) $-2A-B+C=2$ …… (2) $A=7$ …… (3) And put the value of A into equation (1): $\begin{align} & 7+B=1 \\ & B=-6 \end{align}$ Also, put the value of A and B into equation (2): $\begin{align} & -2\left( 7 \right)+6+C=2 \\ & C=2+8 \\ & =10 \end{align}$ Therefore, $\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$ Thus, the partial fraction of the provided expression is $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$.
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