## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 40

#### Answer

The simplified partial fraction expansion is $\frac{3x}{{{x}^{2}}-2x+2}+\frac{x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$.

#### Work Step by Step

The provided rational fraction is as follows: $\frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$ Now, we demonstrate the steps as follows: Step 1: Set up the partial fraction expansion with unknown constant coefficients and then write a constant coefficient over each of the two distinct algebraic linear factors in the denominator of the expression. $\frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}-2x+2}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$ Step 2: Multiply both sides of the equation of the expression by ${{\left( {{x}^{2}}-2x+2 \right)}^{2}}$, considering the least common denominator. ${{\left( {{x}^{2}}-2x+2 \right)}^{2}}\times \left( \frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \right)={{\left( {{x}^{2}}-2x+2 \right)}^{2}}\times \left( \frac{Ax+B}{{{x}^{2}}-2x+2}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \right)$ Then, simplify both sides of the equation as given below: $3{{x}^{3}}-6{{x}^{2}}+7x-2=\left( {{x}^{2}}-2x+2 \right)\left( Ax+B \right)+\left( Cx+D \right)$ \begin{align} & 3{{x}^{3}}-6{{x}^{2}}+7x-2=\left( {{x}^{2}}-2x+2 \right)\left( Ax+B \right)+\left( Cx+D \right) \\ & =A\left( {{x}^{3}}-2{{x}^{2}}+2x \right)+B\left( {{x}^{2}}-2x+2 \right)+Cx+D \\ & =A{{x}^{3}}+{{x}^{2}}\left( -2A+B \right)+x\left( 2A-2B+C \right)+2B+D \end{align} Step 3: And equate the coefficients of like powers of $x$ and of the constant terms. Then we get the system of linear equations with the unknown values of $A$, $B$, $C$ and $D$. $A=3$ (I) $-2A+B=-6$ (II) $2A-2B+C=7$ (III) $2B+D=-2$ (IV) Step 4: Then, solve the system for $A$, $B$, $C$, and $D$. Put the value of $A$ in equation (II) and simplify as follows: \begin{align} & -2A+B=-6 \\ & -2\times 3+B=-6 \\ & B=0 \end{align} Similarly, putting the value of $B$ in equation (IV) and simplifying as follows, we get: \begin{align} & 2B+D=-2 \\ & 2\times 0+D=-2 \\ & D=-2 \end{align} Similarly, putting the value of $A$,$B$, and $D$ in equation (III) and simplifying as follows, we get: \begin{align} & 2A-2B+C=7 \\ & 2\times 3-2\times 0+C=7 \\ & C=7-6 \\ & C=1 \end{align} Step 5: By putting the values of $A$,$B$, $C$ and $D$ and writing the partial function expression, we get: \begin{align} & \frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}-2x+2}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \\ & =\frac{3\times x+0}{{{x}^{2}}-2x+2}+\frac{1\times x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \\ & =\frac{3x}{{{x}^{2}}-2x+2}+\frac{x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \end{align} Thus, $\frac{3x}{{{x}^{2}}-2x+2}+\frac{x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$ is the required partial fraction expansion of the rational expression $\frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$.

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