Having trouble using my usual method of partial fraction decomposition for $\frac{9 + 3s}{s^3 + 2s^2 - s - 2}$. I'm having trouble using my usual method of partial fraction decomposition for $\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2}$.
We can factor such that $$\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2} = \dfrac{A}{s - 1} + \dfrac{B}{s + 1} + \dfrac{C}{s + 2}$$
Therefore, $$ 9 + 3s = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1)$$
And we have that $s \not = 1, -1, -2$.
From here I usually plug in values to find $A$, $B$, and $C$.
$s = 0:$
$$9 = 2A - 2B - C \implies C = 2A - 2B - 9$$
$s = 2:$
$$15 = 12A + 4B + 3C \implies A = \dfrac{15 - 4B - 3C}{12}$$
$s = 3:$
$$18 = 20A + 10B + 8C \implies 10B = 18 - 20A - 8C \implies B = \dfrac{9}{5} - 2A - \dfrac{4C}{5}$$
We now have equations for $A$, $B$, and $C$. But if I try to substitute them into each other, this will result in an infinite loop of substitution.
This method for partial fraction decomposition have always worked for me in the past, so I don't understand why it isn't working in this situation.
I would greatly appreciate it if people could please take the time to explain why my method for partial fraction decomposition is not working in this case and what I should do.

I would prefer a simpler method for this problem, known as Heaviside cover-up method:
$$\dfrac{9+3s}{s^3+2s^2-s-2}=\dfrac{9+3s}{s^2(s+2)-(s+2)}
=\dfrac{9+3s}{(s-1)(s+1)(s+2)}$$
Now,$$\dfrac{9+3s}{(s-1)(s+1)(s+2)}\\=\left.\dfrac{9+3s}{(s+1)(s+2)}\right|_{s=1}\cdot\dfrac{1}{s-1}+\left.\dfrac{9+3s}{(s-1)(s+2)}\right|_{s=-1}\cdot\dfrac{1}{s+1}+\left.\dfrac{9+3s}{(s-1)(s+1)}\right|_{s=-2}\cdot\dfrac{1}{s+2}\\
=\dfrac{2}{s-1}-\dfrac{3}{s+1}+\dfrac{1}{s+2}
$$
