Answer
$\frac{4x^2-x-2}{x^3(x+2)}=\frac{2}{x}+\frac{-1}{x^3}+\frac{-2}{x+2}$,
Work Step by Step
$\frac{4x^2-x-2}{(x^4+2x^3)}$, factorising $x^4+2x^3=x^3(x+2)$:
$\frac{4x^2-x-2}{x^3(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+2}$,
$4x^2-x-2=Ax^2(x+2)+Bx(x+2)+C(x+2)+Dx^3$,
$4x^2-x-2=A(x^3+2x^2)+B(x^2+2x)+C(x+2)+Dx^3$,
$4x^2-x-2=(A+D)x^3+(2A+B)x^2+(2B+C) +2C=4x^2-x-2$
$\begin{cases}
& A & +D= 0&\\
& 2A & +B= 4&\\
& +2B & +C=-1&\\
& 2C=-2
\end{cases}$
thus, $C=-1$,
-substituting $C$ back into the third equation, $2B-1=-1, B=0$.
-substituting $B$ back into the second equation, $2A=4, A=2$.
-substituting $A$ back into the first equation, $2+D=0, D=-2$
Therefore,
$\frac{4x^2-x-2}{x^3(x+2)}=\frac{2}{x}+\frac{-1}{x^3}+\frac{-2}{x+2}$,
