Answer
$\frac{6}{x+4}-\frac{3}{x-1}-\frac{1}{x+2}$
Work Step by Step
1. Factor the denominator to get $(x-1)(x^2+6x+8)=(x-1)(x+2)(x+4)$
2. Assume $\frac{2x^2-15x-32}{(x-1)(x^2+6x+8)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x+4}$
3. We have $RHS=\frac{A(x+2)(x+4)+B(x-1)(x+4)+C(x-1)(x+2)}{(x-1)(x+2)(x+4)}=\frac{(A+B+C)x^2+(6A+3B+C)x+8A-4B-2C}{(x-1)(x+2)(x+4)}$
4. Compare steps 2 and 3 to get $\begin{cases} A+B+C=2 \\ 6A+3B+C=-15 \\8A-4B-2C=-32 \end{cases}$ or $\begin{cases} A+B+C=2 \\ 6A+3B+C=-15 \\4A-2B-C=-16 \end{cases}$
5. Add the third equation to the first and the second to get $\begin{cases} 5A-B=-14 \\ 10A+B=-31 \end{cases}$
6. Add up the above two equations to get $15A=-45$, thus $A=-3$
7. We have $B=5A+14=-1$ and $C=2-A-B=6$
8, Thus we have $\frac{2x^2-15x-32}{(x-1)(x^2+6x+8)}=\frac{-3}{x-1}+\frac{-1}{x+2}+\frac{6}{x+4}=\frac{6}{x+4}-\frac{3}{x-1}-\frac{1}{x+2}$
