Answer
$\frac{3x^2-2x+8}{x^3-x^2+2x-2}=\frac{3}{x-1}+\frac{-2}{x^2+1}$
Work Step by Step
$\frac{3x^2-2x+8}{x^3-x^2+2x-2}$, factorising $x^3-x^2+2x-2=x^2(x-1)+2(x-1)=(x^2+2)(x-1)$,
$\frac{3x^2-2x+8}{x^3-x^2+2x-2}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$,
$A(x^2+2)+Bx+C(x-1)$,
$Ax^2+A+Bx^2-Bx+Cx-C$,
$(A+B)x^2+(-B+C)x+(A-C)=3x^2-2x+8$,
$\begin{cases}
A+B=3\\
-B+C=-2\\
2A-C=8
\end{cases}$
Adding Equation 2 and Equation 3.
$\begin{cases}
-B+C=-2\\
2A-C=8\\
-- -- -- --\\
2A-B=6
\end{cases}$
Adding Equation 1 and the addition result of Equation 2 and Equation 3.
$\begin{cases}
A+B=3\\
2A-B=6\\
-- -- -- --\\
3A=9
\end{cases}$
thus, $A=3$, substituting back into the equation $3+B=3, B=0$
substituting back into the equation $6-C=8, C=-2$.
Therefore,
$\frac{3x^2-2x+8}{x^3-x^2+2x-2}=\frac{3}{x-1}+\frac{-2}{x^2+1}$
