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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.

Damian Taylor
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2 Answers2

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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!).

Robert Z
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  • How do you proceed with the polynomial division? – Damian Taylor Mar 28 '18 at 15:18
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    In this case the have the same degree and the same principal term so just subtract the denominator to the numerator. In general see https://en.wikipedia.org/wiki/Polynomial_long_division – Robert Z Mar 28 '18 at 15:20
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for your control: it is $$1+2\, \left( x+2 \right) ^{-1}- \left( x+5 \right) ^{-1}-3\, \left( x- 1 \right) ^{-1} $$

Dr. Sonnhard Graubner
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  • It seems (based on comparing your result to the one in the OP) that the OP may simply have gotten $a,b,c$ rearranged. It would certainly help to give details for how you arrived at the above. – hardmath Mar 28 '18 at 17:24