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: 26


The partial fraction is, $\frac{1}{x}+\frac{2}{\left( x+7 \right)}-\frac{28}{{{\left( x+7 \right)}^{2}}}$

Work Step by Step

$\frac{3{{x}^{2}}+49}{x{{\left( x+7 \right)}^{2}}}=\frac{A}{x}+\frac{B}{\left( x+7 \right)}+\frac{C}{{{\left( x+7 \right)}^{2}}}$ Now, multiply both sides by $x{{\left( x+7 \right)}^{2}}$; then we get, $\begin{align} & 3{{x}^{2}}+49=A{{\left( x+7 \right)}^{2}}+Bx\left( x+7 \right)+Cx \\ & =A\left( {{x}^{2}}+49+14x \right)+B{{x}^{2}}+7Bx+Cx \\ & =A{{x}^{2}}+49A+14Ax+B{{x}^{2}}+7Bx+Cx \\ & =\left( A+B \right){{x}^{2}}+\left( 14A+7B+C \right)x+49A \end{align}$ Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+B=3$ …… (1) $14A+7B+C=0$ …… (2) $49A=49$ …… (3) Since, $49A=49$. It implies: $A=1$ Put $A=1$ in equation (1) to get: $\begin{align} & 1+B=3 \\ & B=3-1 \\ & =2 \end{align}$ It implies $B=2$ Now, put $A=1$ and $B=2$ in equation (2); then we get: $\begin{align} & 14\left( 1 \right)+7\left( 2 \right)+C=0 \\ & 28+C=0 \\ & C=-28 \end{align}$ Therefore, $\frac{3{{x}^{2}}+49}{x{{\left( x+7 \right)}^{2}}}=\frac{1}{x}+\frac{2}{\left( x+7 \right)}-\frac{28}{{{\left( x+7 \right)}^{2}}}$ Thus, the partial fraction of the provided expression is $\frac{1}{x}+\frac{2}{\left( x+7 \right)}-\frac{28}{{{\left( x+7 \right)}^{2}}}$.
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