<s>[INST] Question about partial fractions with irreducible quadratic factors Given this rational function: $$\frac{-4x^4-2x^3-26x^2-8x-44}{(x+1)(x^2 +3)^2}$$
The decomposition would look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
And the final answer would be: $$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
But, if you were to set it up like this: $$\frac{A}{x+1} + \frac{B}{(x^2+3)} + \frac{C}{(x^2+3)^2}$$
You end up with the same answer:
$$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
I want to know why this is. 
I thought that for irreducible quadratic factors, you needed a linear term in the numerator. Is this purely coincidence that it worked with constant terms in the numerator?
 [/INST] The five functions 
$$f_1(x)=\frac{1}{x+1},\; f_2(x)=\frac{x}{x^2+3}, \; f_3(x)=\frac{1}{x^2+3},\; f_4(x)=\frac{x}{(x^2+3)^2}, \; f_5(x)=\frac{1}{(x^2+3)^2}.$$
is a basis which linearly generates all rational functions of the form
$$\frac{P(x)}{(x+1)(x^2 +3)^2}$$
where $P$ is ANY polynomial of degree $\leq 4$.
In your case, we simply have that in the vector representation of the given function the coordinates related to $f_2$ and $f_4$ are zero. So, yes, this is a purely coincidence.
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