Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of 
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x$
No clue on my next step or even if this is the right step.
 A: Compare coefficients of both sides polynomials. for example by comparing coefficient next you $x^2$ you will get $A+B+C = 3$. Do the same for $x^1$ and $x^0$ and solve system of 3 linear equations.
The other way is to put arbitrary 3 values of $x$, for example $-1$, $0$ and $1$, and again you will get system of 3 linear aquations with variables $A$, $B$ and $C$.
A: $$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x.$$
Setting $x=-4$,
$$0A+0B+18C=-20.$$
A: $$
\begin{align*}
\left(A+B+C\right) x^2 &= 3x^2 \\
\left(2A + 5B -C\right)x &= 17x \\
-8A + 4B - 2C &= 0
\end{align*}
$$  
Dividing the first equation by $x^2$ and the second by $x$ will get you a linear system of equations.
A: Hint
You have $$\frac{3 x^2+17 x}{x^3+3 x^2-6 x-8}=\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$$ So, write $$f(x)=3x^2+17x-\Big(A(x-2)(x+4)+B(x+1)(x+4)+C(x+1)(x-2)\Big)=0$$ Now $$f(2)=3\times 2^2+17 \times 2-B\times(2+1)\times(2+4)=46-18B=0$$ Repeat for $f(-1)$ and $f(-4)$; you just have one linear equation at the time.
A: if you want to find just $C$ not other two, then look at the behavior of $$ \frac{3x^2 + 17x}{(x+1)(x-2)(x+4)} = \frac{3*(-4)^2+17*(-4)}{(-4+1)(-4-2)(x+4)} +\cdots=-\frac{10/9}{(x-4)}+\cdots$$  therefore $$C = -\frac{10}9. $$
