Answer
$\dfrac{9x+2}{(x-2)(x^2+2x+2)}=\dfrac{2}{x-2}+\dfrac{-2x+1}{x^2+2x+2}$
Work Step by Step
We are given the fraction:
$\dfrac{9x+2}{(x-2)(x^2+2x+2)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{9x+2}{(x-2)(x^2+2x+2)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+2x+2}$
Multiply all terms by the least common denominator $(x-2)(x^2+2x+2)$:
$(x-2)(x^2+2x+2)\cdot\dfrac{9x+2}{(x-2)(x^2+2x+2)}=(x-2)(x^2+2x+2)\cdot\dfrac{A}{x-2}+(x-2)(x^2+2x+2)\cdot\dfrac{Bx+C}{x^2+2x+2}$
$9x+2=A(x^2+2x+2)+(Bx+C)(x-2)$
$9x+2=Ax^2+2Ax+2A+Bx^2-2Bx+Cx-2C$
$9x+2=(A+B)x^2+(2A-2B+C)x+(2A-2C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=0\\
2A-2B+C=9\\
2A-2C=2
\end{cases}$
$\begin{cases}
A+B=0\\
2A-2B+C=9\\
A-C=1
\end{cases}$
Solve the system: add Equation 3 to Equation 2:
$\begin{cases}
A+B=0\\
2A-2B+C+A-C=9+1
\end{cases}$
$\begin{cases}
A+B=0\\
3A-2B=10
\end{cases}$
$\begin{cases}
2A+2B=2(0)\\
3A-2B=10
\end{cases}$
$2A+2B+3A-2B=0+10$
$5A=10$
$A=2$
$A+B=0$
$2+B=0$
$B=-2$
$A-C=1$
$2-C=1$
$C=1$
The partial fraction decomposition is:
$\dfrac{9x+2}{(x-2)(x^2+2x+2)}=\dfrac{2}{x-2}+\dfrac{-2x+1}{x^2+2x+2}$
