## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{2}{\left( x-2 \right)}+\frac{-2x+1}{\left( {{x}^{2}}+2x+2 \right)}$.
$\frac{9x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A}{\left( x-2 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+2x+2 \right)}$. Now, take the L.C.M on the right side: $\frac{9x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x-2 \right)}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}$. By eliminating the denominators from both sides, we get: \begin{align} & 9x+2=A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x-2 \right) \\ & =A{{x}^{2}}+2Ax+2A+B{{x}^{2}}-2Bx+Cx-2C \\ & =\left( A+B \right){{x}^{2}}+\left( 2A+C-2B \right)x+2A-2C \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+B=0$ …… (I) $2A+C-2B=9$ …… (II) $2A-2C=2$ $A-C=1$ …… (III) Now, find the values of A and B from equation (III) and equation (I) respectively and put in equation (II): \begin{align} & A=1+C \\ & B=-A \\ & B=-C-1 \\ \end{align} \begin{align} & 2\left( 1+C \right)+C-2\left( -C-1 \right)=9 \\ & 2+2C+C+2C+2=9 \\ & 5C=5 \\ & C=1 \end{align} Then, put the value of C in equation (III): \begin{align} & A-1=1 \\ & A=2 \end{align} And put the value of A in equation (I): \begin{align} & 2+B=0 \\ & B=-2 \end{align} Then, $\frac{9x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{2}{\left( x-2 \right)}+\frac{-2x+1}{\left( {{x}^{2}}+2x+2 \right)}$ Thus, the partial fraction of the provided expression is $\frac{2}{\left( x-2 \right)}+\frac{-2x+1}{\left( {{x}^{2}}+2x+2 \right)}$.