Answer
$\dfrac{3x^3-6x^2+7x-2}{(x^2-2x+2)^2}=\dfrac{3x}{x^2-2x+2}+\dfrac{x-2}{(x^2-2x+2)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{3x^3-6x^2+7x-2}{(x^2-2x+2)^2}$
As the denominator is factored, we can write the partial fraction decomposition:
$\dfrac{3x^3-6x^2+7x-2}{(x^2-2x+2)^2}=\dfrac{Ax+B}{x^2-2x+2}+\dfrac{Cx+D}{(x^2-2x+2)^2}$
Multiply all terms by the least common denominator $(x^2-2x+2)^2$:
$(x^2-2x+2)^2\cdot\dfrac{3x^3-6x^2+7x-2}{(x^2-2x+2)^2}=(x^2-2x+2)^2\cdot\dfrac{Ax+B}{x^2-2x+2}+(x^2-2x+2)^2\cdot\dfrac{Cx+D}{(x^2-2x+2)^2}$
$3x^3-6x^2+7x-2=(Ax+B)(x^2-2x+2)+(Cx+D)$
$3x^3-6x^2+7x-2=Ax^3-2Ax^2+2Ax+Bx^2-2Bx+2B+Cx+D$
$3x^3-6x^2+7x-2=Ax^3+(-2A+B)x^2+(2A-2B+C)x+(2B+D)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A=3\\
-2A+B=-6\\
2A-2B+C=7\\
2B+D=-2
\end{cases}$
Solve the system:
$A=3$
$\begin{cases}
-2(3)+B=-6\\
2(3)-2B+C=7\\
2B+D=-2
\end{cases}$
$\begin{cases}
-6+B=-6\\
6-2B+C=7\\
2B+D=-2
\end{cases}$
$B=0$
$\begin{cases}
6-2(0)+C=7\\
2(0)+D=-2
\end{cases}$
$\begin{cases}
6+C=7\\
D=-2
\end{cases}$
$C=1$
$D=-2$
The partial fraction decomposition is:
$\dfrac{3x^3-6x^2+7x-2}{(x^2-2x+2)^2}=\dfrac{3x}{x^2-2x+2}+\dfrac{x-2}{(x^2-2x+2)^2}$
