## Precalculus (6th Edition) Blitzer

The partial fraction decomposition is ${{x}^{3}}+4{{x}^{2}}+12x+32+\frac{80}{x-2}+\frac{32}{{{\left( x-2 \right)}^{2}}}$.
Let us consider the expression, $\frac{{{x}^{5}}}{{{x}^{2}}-4x+4}$ Firstly, solve by long division method, {{x}^{2}}-4x+4\overset{{{x}^{3}}+4{{x}^{2}}+12x+32}{\overline{\left){\begin{align} & {{x}^{5}} \\ & {{x}^{5}}\text{ }-\text{4}{{x}^{4}}+4{{x}^{3}} \\ & -\text{ }+\text{ }- \\ & \overline{\begin{align} & \text{ 4}{{x}^{4}}-4{{x}^{3}} \\ & \text{ 4}{{x}^{4}}-\text{16}{{x}^{3}}+16{{x}^{2}} \\ & \text{ }-\text{ + }- \\ & \overline{\begin{align} & \text{ 12}{{x}^{3}}-16{{x}^{2}} \\ & \text{ 12}{{x}^{3}}-48{{x}^{2}}+48x \\ & \text{ + }- \\ & \overline{\begin{align} & \text{ 32}{{x}^{2}}-48x \\ & \text{ 32}{{x}^{2}}-128x+128 \\ & \text{ }-\text{ + }- \\ & \overline{\text{ 80}x-128} \\ \end{align}} \\ \end{align}} \\ \end{align}} \\ \end{align}}\right.}} So, $\frac{{{x}^{5}}}{{{x}^{2}}-4x+4}={{x}^{3}}+4{{x}^{2}}+12x+32+\frac{80x-128}{{{\left( x-2 \right)}^{2}}}$ Then, partial fraction of $\frac{80x-128}{{{x}^{2}}-4x+4}=\frac{80x-128}{{{\left( x-2 \right)}^{2}}}$ And to find partial fraction decomposition, $\frac{80x-128}{{{\left( x-2 \right)}^{2}}}=\frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}}$ …… (1) And multiply both sides by ${{\left( x-2 \right)}^{2}}$, \begin{align} & 80x-128=A\left( x-2 \right)+B \\ & 80x-128=Ax+2A+B \\ \end{align} And equate the terms, \begin{align} & A=80 \\ & -2A+B=-128 \end{align} Then, solving the above equation, $A=80,B=32$ And put these values in equation (1), $\frac{80x-128}{{{\left( x-2 \right)}^{2}}}=\frac{80}{x-2}+\frac{32}{{{\left( x-2 \right)}^{2}}}$ Thus, the partial fraction decomposition of the remainder term is, ${{x}^{3}}+4{{x}^{2}}+12x+32+\frac{80}{x-2}+\frac{32}{{{\left( x-2 \right)}^{2}}}$.