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.

You've got the right expression
$$
9 + 3s = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1).
$$
Now you can think like this: it is a polynomial equation of the kind $p(s)=0$. A nonzero polynomial cannot have more than the finite number of zeros. Since this equation has infinitely many solutions (all $s$ except a finite number) then the polynomial must be identical zero, and the equation holds true for those exceptional values as well. Setting $s=\pm 1, -2$ we get $A=2$, $B=-3$, $C=1$.
It is also possible - as you did - to set other values of $s$ and get the linear system. In your case you have obtained
$$
\begin{cases}
2A-2B-C=9,\\
12A+4B+3C=15,\\
20A+10B+8C=18.
\end{cases}
$$
Using the linear algebra technique such systems are usually solved by Gauss elimination, but it is also possible to solve it by substitution as you tried, but got infinite loops. You have to do a systematic substitution: express one variable e.g. $C$ from one equation and substitute it into all other equations. Here how it goes: the first equation gives $C=2A-2B-9$. Substitute it into the second and the third ones
$$
\begin{cases}
C=2A-2B-9,\\
12A+4B+3(2A-2B-9)=15,\\
20A+10B+8(2A-2B-9)=18.
\end{cases}\quad\Leftrightarrow\quad
\begin{cases}
C=2A-2B-9,\\
9A-B=21,\\
18A-3B=45.
\end{cases}
$$
Now we do the same: express $B$ from the second equation $B=9A-21$ and set into the third one
$$
\begin{cases}
C=2A-2B-9,\\
B=9A-21,\\
18A-3(9A-21)=45.
\end{cases}\quad\Leftrightarrow\quad
\begin{cases}
C=2A-2B-9,\\
B=9A-21,\\
A=2.
\end{cases}
$$
It works without infinite loops.
