<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] In the first expansion B = D = 0, but with a different but similar question that would not necessarily happen.
I can provide an example easily by starting with values for A = B = C = D = E = 1, working backwards to a new question that will not have the coincidental property that your one has.
Here is the new question:
$$\frac{2x^4+2x^3+11x^2+8x+44}{(x+1)(x^2 +3)^2}$$
The decomposition will, as before, look like this:
$$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
and with the values substituted in,
$$\frac{1}{x+1} + \frac{x+1}{(x^2+3)} + \frac{x+1}{(x^2+3)^2}$$
Robert Z has added a neat answer that goes further and looks at the underlying theory.
Incidentally, the free online Wolfram Alpha is really handy for playing around with these sort of expressions and it'll let you check your answers from partial fraction questions.
Link  https://www.wolframalpha.com/
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