Crawfish

I just typed out how to do it and then my post was deleted 

I'm just gonna give the summary:

fill in a,b,c,d for the blanks.

Then, find a common denominator for all variables on the right side. At this point it should look like the image below:

$\frac{8x^3 + 24x^2 + 15x + 1}{(x+1)(x-1)(x)(x+3)} = \frac{a(x+1)x(x+3)}{(x+1)(x-1)(x)(x+3)} + \frac{b(x+1)x(x-1)}{(x+1)(x-1)(x)(x+3)} + \frac{c(x+1)(x-1)(x+3)}{(x+1)(x-1)(x)(x+3)} + \frac{d(x-1)x(x+3)}{(x+1)(x-1)(x)(x+3)}$

You can cancel the denominator on both sides and foil the polynomials on the right side (but do not multiply by a,b,c or d):

 $8x^3 + 24x^2 + 15x + 1 = a(x^3 + 4x^2 + 3x) + b(x^3 + x^2 - x - 1) + c(x^3 + 2x^2 - 3x) + d(x^3 - 2x^2 + 1)$

$x^3$ can only be paired with $x^3$, $x^2$ with $x^2$, and so on. This means $ax^3+bx^3+cx^3+dx^3=8x^3$, or $a+b+c+d=8$. We can apply this rule to $x^3, x^2, x,$ and the constant. So we get:

$\begin{aligned} a + b + c + d &= 8 \\ 4a + b + 2c - 3d &= 24 \\ 3a - b - 3c &= 15 \\ -c + d &= 1 \end{aligned}$

You may use any method to solve this system of equations. The first step is to substitute $d=c+1$ into every equation. I used matrices to solve from there, but you may want to add/subtract equations depending on your knowledge. You should get the final answer:

p.s: you would probably have more success in getting replies if you typed your equation into the website's built-in LaTeX editor. Just sayin'