## Precalculus (6th Edition) Blitzer

The partial fraction decomposition is ${{x}^{2}}+3x+1+\frac{3}{x+1}+\frac{5}{x-2}$.
Let us consider the expression, $\frac{{{x}^{4}}+2{{x}^{3}}-4{{x}^{2}}+x-3}{{{x}^{2}}-x-2}$ Firstly, solve by long division method, {{x}^{2}}-x-2\overset{{{x}^{2}}+3x+1}{\overline{\left){\begin{align} & {{x}^{4}}+2{{x}^{3}}-4{{x}^{2}}+x-3 \\ & {{x}^{4}}-{{x}^{3}}\text{ }-2{{x}^{2}} \\ & -\text{ + +} \\ & \overline{\begin{align} & \text{ 3}{{x}^{3}}\text{ }-\text{2}{{x}^{2}}+x-3 \\ & \text{ 3}{{x}^{3}}\text{ }-3{{x}^{2}}-\text{6}x \\ & \text{ }-\text{ + +} \\ & \overline{\begin{align} & \text{ }{{x}^{2}}\text{ +7}x-3 \\ & \text{ }{{x}^{2}}\text{ }-x\text{ }-\text{2} \\ & \text{ }-\text{ + +} \\ & \overline{\text{ 8}x-1} \\ \end{align}} \\ \end{align}} \\ \end{align}}\right.}} So, $\frac{{{x}^{4}}+2{{x}^{3}}-4{{x}^{2}}+x-3}{{{x}^{2}}-x-2}={{x}^{2}}+3x+1+\frac{8x-1}{{{x}^{2}}-x-2}$ Then, partial fraction of $\frac{8x-1}{{{x}^{2}}-x-2}=\frac{8x-1}{\left( x+1 \right)\left( x-2 \right)}$, And to find partial fraction decomposition, $\frac{8x-1}{\left( x+1 \right)\left( x-2 \right)}=\frac{A}{x+1}+\frac{B}{x-2}$ …… (1) And multiply both sides by $\left( x+1 \right)\left( x-2 \right)$, \begin{align} & 8x-1=A\left( x-2 \right)+B\left( x+1 \right) \\ & 8x-1=Ax-2A+Bx+B \\ & 8x-1=\left( A+B \right)x-\left( 2A-B \right) \\ \end{align} And equate the terms, \begin{align} & A+B=8 \\ & -2A+B=-1 \\ \end{align} Then solving above equation, $A=3,B=5$ Put these values in equation (1), $\frac{8x-1}{\left( x+1 \right)\left( x-2 \right)}=\frac{3}{x+1}+\frac{5}{x-2}$ Thus, the partial fraction decomposition of the remainder term is, ${{x}^{2}}+3x+1+\frac{3}{x+1}+\frac{5}{x-2}$.