Partial Fraction Solution? This is the original fraction : $$\dfrac{8s^2}{(2s-3)^3}$$
From this I get : 
$$4As^2 - 12As + 9A + 2Bs - 3B + C = 8s^2$$
On the solution it says $4A = 8, A=2$, okay I understand this, $4As^2 = 8s^2$, and the rest is cancelled in the equation since there is only one $s^2$, but how $B=12$ and $C = 18$, I don't understand, can somebody explain please?
 A: $$\begin{align} -12 As + 2 Bs = 0s &\implies -12 A + 2B = 0 \\ \\ &\iff -12(2) + 2B = 0 \tag{A = 2}\\ \\ &\iff 2B = 24 \\ \\ &\iff B = 12\tag{B = 12}\end{align}$$
Do likewise for the constant term: $$9A - 3B + C = 0$$ Substitute into the equation your values $A = 2,\; B= 12$ to solve  for $C$.
A: As alternative to expanding polynomial expression and equating coefficients as you have done, you can exploit the fact that two polynomials are equal iff they have the same values for all possible values of the variable. So starting with $8s^2 = A(2s-3)^2 + B(2s-3) + C$, we can choose values for $s$ and equate both sides. This can be a much faster method for partial fractions with lots of factors in the denominator. For your problem, I'd pick $s=\frac{3}{2}, 1, 2$ and evaluate them in that order. The reason for those numbers is that $\frac{3}{2}$ is the solution to the linear factors present so it makes for easy evaluation. The other two are chosen because they yield easy to work with results from $2s-3$.
$s=\frac{3}{2} \Longrightarrow 8(\frac{3}{2})^2 = 18 = A(0)^2 + B(0)^2 +C \Longleftrightarrow C = 18$
$s = 1 \Longrightarrow 8(1)^2 = A(-1)^2 + B(-1) + C \Longleftrightarrow 8 = A - B + 18 \Longleftrightarrow A - B = -10$
$s = 2 \Longrightarrow 8(2)^2 = A(1)^2 + B(1) + C \Longleftrightarrow 32 = A + B + 18 \Longleftrightarrow A + B = 14$ 
A very quick solve the system of equation generated from $s=1$ and $s=2$ yields that $A = 2$ and $B=12$.
