## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 24

#### Answer

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

#### Work Step by Step

$\frac{2{{x}^{2}}+8x+3}{{{\left( x+1 \right)}^{3}}}=\frac{A}{\left( x+1 \right)}+\frac{B}{{{\left( x+1 \right)}^{2}}}+\frac{C}{{{\left( x+1 \right)}^{3}}}$ Now, multiply both sides by ${{\left( x+1 \right)}^{2}}$; then we get \begin{align} & 2{{x}^{2}}+8x+3=A{{\left( x+1 \right)}^{2}}+B\left( x+1 \right)+C \\ & =A\left( {{x}^{2}}+1+2x \right)+Bx+B+C \\ & =A{{x}^{2}}+A+2Ax+Bx+B+C \\ & =A{{x}^{2}}+\left( 2A+B \right)x+A+B+C \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A=2$ …… (1) $2A+B=8$ …… (2) $A+B+C=3$ …… (3) And put the value of A in equation (2): \begin{align} & 2\left( 2 \right)+B=8 \\ & B=8-4 \end{align} $B=4$ …… (4) Also, put the value of A and B in equation (3): \begin{align} & 2+4+C=3 \\ & C=3-6 \\ & =-3 \end{align} Therefore, $\frac{2{{x}^{2}}+8x+3}{{{\left( x+1 \right)}^{3}}}=\frac{2}{\left( x+1 \right)}+\frac{4}{{{\left( x+1 \right)}^{2}}}-\frac{3}{{{\left( x+1 \right)}^{3}}}$ Thus, the partial fraction of the provided expression is $\frac{2}{\left( x+1 \right)}+\frac{4}{{{\left( x+1 \right)}^{2}}}-\frac{3}{{{\left( x+1 \right)}^{3}}}$.

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