## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Review Exercises - Page 877: 20

#### Answer

The partial fraction decomposition of the rational expression is $-\frac{4}{\left( x-1 \right)}+\frac{4}{x-2}-\frac{2}{{{\left( x-2 \right)}^{2}}}$

#### Work Step by Step

We know that the linear factor $x-2$ occurs multiple times in the denominator of the rational expression; therefore, for each power of $x-2$, assign an undefined constant factor in the denominator by the method of grouping as shown below: $\frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\frac{A}{\left( x-1 \right)}+\frac{B}{x-2}+\frac{C}{{{\left( x-2 \right)}^{2}}}$ Multiply both sides of the equation by $\left( x-1 \right){{\left( x-2 \right)}^{2}}$: \begin{align} & \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\left( x-1 \right){{\left( x-2 \right)}^{2}}\left( \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{{{\left( x-2 \right)}^{2}}} \right) \\ & \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\left\{ \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{A}{x-1}+\left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{B}{x-2}+\left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{C}{{{\left( x-2 \right)}^{2}}} \right\} \end{align} And divide the common factors: \begin{align} & \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\left\{ \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{A}{x-1}+\left( x-1 \right)\left( x-2 \right)\left( x-2 \right)\cdot \frac{B}{x-2}+\left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{C}{{{\left( x-2 \right)}^{2}}} \right\} \\ & 2x-6={{\left( x-2 \right)}^{2}}\cdot A+\left( x-1 \right)\left( x-2 \right)\cdot B+\left( x-1 \right)C \\ & 2x-6=\left( {{x}^{2}}+4-4x \right)A+\left( {{x}^{2}}-2x-x+2 \right)B+\left( x-1 \right)C \\ & 2x-6=\left( A+B \right){{x}^{2}}+\left( -4A-3B+C \right)x+\left( 4A+2B-C \right) \end{align} Then, equate the coefficients of like terms of the equation to write a system of equations as given below: $A+B=0$ (I) $-4A-3B+C=2$ (II) $4A+2B-C=-6$ (III) From equation (I), it is inferred that $A=-B$. Put $A=-B$ in equation (II) to obtain the equation: \begin{align} & -4A-3B+C=2 \\ & -4\left( -B \right)-3B+C=2 \\ & 4B-3B+C=2 \\ & B+C=2 \end{align} (IV) Put $A=-B$ in equation (III) to obtain the equation: \begin{align} & 4A+2B-C=-6 \\ & 4\left( -B \right)+2B-C=-6 \\ & -4B+2B-C=-6 \\ & -2B-C=-6 \end{align} (V) Add equation (IV) with equation (V) to get the value of B: \begin{align} & +\text{ }\underline{\begin{align} & B+C=2 \\ & -2B-C=-6 \end{align}} \\ & \text{ }-B=-4 \\ \end{align} Multiply both sides by $-1$: \begin{align} & -1\left( -B \right)=\left( -4 \right)\left( -1 \right) \\ & B=4 \end{align} Therefore, the value of B is $4$. Now, substitute the value of B in equation (IV) to get the value of B: \begin{align} & B+C=2 \\ & 4+C=2 \\ & C=2-4 \\ & C=-2 \end{align} As $A=-B$, therefore, $A=-4$ Put the values of $A,\,\,B,\,\,\text{and }C$ in the given equation, and determine the partial fraction decomposition: \begin{align} & \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\frac{A}{\left( x-1 \right)}+\frac{B}{x-2}+\frac{C}{{{\left( x-2 \right)}^{2}}} \\ & =\frac{-4}{\left( x-1 \right)}+\frac{4}{x-2}+\frac{-2}{{{\left( x-2 \right)}^{2}}} \\ & =-\frac{4}{\left( x-1 \right)}+\frac{4}{x-2}-\frac{2}{{{\left( x-2 \right)}^{2}}} \end{align} Thus, the partial fraction decomposition of the rational expression is $-\frac{4}{\left( x-1 \right)}+\frac{4}{x-2}-\frac{2}{{{\left( x-2 \right)}^{2}}}$.

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