## Precalculus (6th Edition) Blitzer

The partial fraction of the expression is $\frac{-1}{x-2}+\frac{3x-2}{{{x}^{2}}+2x+2}$.
If an expression is of the form $\frac{p{{x}^{2}}+qx+r}{\left( x-a \right)\left( {{x}^{2}}+bx+c \right)}$ then its partial fraction expression is $\frac{A}{x-a}+\frac{Bx+c}{{{x}^{2}}+bx+c}$ Then, rewrite the fraction as: \begin{align} & \frac{2{{x}^{2}}-10x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A}{x-2}+\frac{Bx+C}{{{x}^{2}}+2x+2} \\ & =\frac{A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x-2 \right)}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)} \end{align} $2{{x}^{2}}-10x+2=A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x-2 \right)$ (I) Put $x=2$ in (I), \begin{align} & 2{{\left( 2 \right)}^{2}}-10\left( 2 \right)+2=A\left( {{2}^{2}}+2\cdot 2+2 \right)+0 \\ & 8-20+2=A\left( 4+4+2 \right) \\ & -10=A\left( 10 \right) \\ & A=-1 \end{align} Again, put $x=0$ in (I), \begin{align} & 2=A\left( 2 \right)+C\left( -2 \right) \\ & 1=A-C \\ & C=A-1 \end{align} And simplify it further to get: \begin{align} & C=-1-1 \\ & C=-2 \\ \end{align} Again, put $x=1$ in (I), \begin{align} & 2-10+2=A\left( 1+2+2 \right)+\left( B+C \right)\left( -1 \right) \\ & -6=-1\left( 5 \right)+\left( B-2 \right)\left( -1 \right) \\ & -6=-5-B+2 \end{align} And simplify it further to get \begin{align} & B=6-5+2 \\ & B=3 \\ \end{align} Therefore, $\frac{2{{x}^{2}}-10x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{-1}{x-2}+\frac{3x-2}{{{x}^{2}}+2x+2}$. Thus, the partial fraction is $\frac{-1}{x-2}+\frac{3x-2}{{{x}^{2}}+2x+2}$.