Partial fractions when numerator polynomial has greater degree than denominator

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?

• Then you do division with rest to the numerator. Commented Jan 3, 2021 at 23:21
• Yes. In this case, the canonical way uses Euclidean division of the numerator by the denominator, so that it comes down to a proper rational function. Commented Jan 3, 2021 at 23:32

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.