Answer
The partial fraction of the expression is $\frac{-1}{x-2}+\frac{3x-2}{{{x}^{2}}+2x+2}$.
Work Step by Step
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 $\frac{-1}{x-2}+\frac{3x-2}{{{x}^{2}}+2x+2}$.
