Consider the 'The Big example' as shown in this site, in it we are tasked to split:
$$ \frac{x^2 +15}{(x^2+3)(x+3)^2} $$
They split it as:
$$ \frac{A_1}{x+3} + \frac{A_2}{(x+3)^2} + \frac{Bx+C}{(x^2+3)}$$
Suppose, the degree of the numerator polynomial was greater than degree of expanded polynomial in numeartor. For example say it was $x^7+15$, it seems clear to me that the above split fails. So, does there exist a split which is applicable even in such cases?
If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division to obtain $$P(x)+\frac{A_1}{x+3}+\frac{A_2}{(x+3)^2}+\frac{Bx+C}{x^2+3},$$ where $P(x)$ is a polynomial.
This expansion results in $$A_1(x^2+3)(x+3)+A_2(x^2+3)+(Bx+C)(x+3)^2=x^2+15$$ $$\implies(A_1+B)x^3+(3A_1+A_2+6B+C)x^2+(3A_1+9B+6C)x+(9A_1+3A_2+9C)=x^2+15$$
Forcing through $x=-3\implies A_2=2$, we create a system: $$\begin{array}{lcr} A_1 & B & C & |& P \\\hline 1 & 1 & 0 &|& 0 \\ 3 & 6 & 1 & |&-1 \\ 3 & 9 & 6 &|& 0 \\ 9 & 0 & 9 & |&9 \end{array}$$ which Gaussian elimination handles. Notice that my left-hand polynomial has an $x^3$ and $x$ term, but the right-hand does not, so I just consider them as $0$ and solve normally.
