Partial fraction decomposition with a 3rd degree numerator

Question

I have the following function to decompose using PFD:

$H(x) = \frac{x^3 + 4x^2 - 11x - 48}{x^3 + 6x^2 + 3x - 10}$

The poles are $1$, $-2$ and $-5$ so I tried to do it this way: finding $a$, $b$ and $c$ such as:

$H(x) = \frac{a}{x-1} + \frac{b}{x+2} + \frac{c}{x+5}$

But it didn't give me a correct result. Doing it that way gives me $(a,b,c) = (-3, 2, -1) $

Is there a specific form of numerator to introduce here? Thanks.

Answer 1

Since the degree of the numerator is $\geq$ to the degree of the denominator you should first divide them. Hence $$\frac{x^3 + 4x^2 - 11x - 48}{x^3 + 6x^2 + 3x - 10}=1-\frac{2x^2 + 14x + 38}{x^3 + 6x^2 + 3x - 10}=1+\frac{a}{x-1} + \frac{b}{x+2} + \frac{c}{x+5}$$ for some real constants $a$, $b$, $c$ (the same that you already found by using the residue method!).

Answer 2

for your control: it is $$1+2\, \left( x+2 \right) ^{-1}- \left( x+5 \right) ^{-1}-3\, \left( x- 1 \right) ^{-1} $$