<s>[INST] Express ${x^3 + 2x^2 + 61 \over (x + 3)^2 (x^2 + 4)}$ in partial fractions $$\eqalign{
  & {x^3 + 2x^2 + 61 \over (x + 3)^2(x^2 + 4)} \equiv {A \over (x + 3)} + {B \over (x + 3)^2} + {Cx + D \over (x^2 + 4)}  \cr 
  &  \equiv {A(x + 3)(x^2 + 4) + B(x^2 + 4) + (Cx + D)(x + 3)^2 \over (x + 3)^2(x^2 + 4)} \cr} $$
so:
$$x^3 + 2x^2 + 61 \equiv A(x + 3)(x^2 + 4) + B(x^2 + 4) + (Cx + D)(x + 3)^2$$
Solving for $B$, making $x=-3$:
$$\begin{align}
 - 27 + 18 + 61 & = 0 + 13B + 0  \\[6pt]
  13B & = 52  \\[6pt]
  B & = 4 \end{align} $$
The other three variables are tricky as there is no value that I can see that is capable of eliminating two and leaving one, I've tried to "compare coefficients" but I am left with more than one unknown so I think the only other options available to me is solving simultaneously with 3 variables (something I do not know how to do), have i missed something that could make this easier to write in partial fractions?
Thank you.
 [/INST] You're doing the problem right.  I don't think there's a way of avoiding solving a $3\times 3$ system.  However, you can try "nice" values for $x$ to make it a nice system, e.g. $x=0, x=1, x=-1$.
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