Why a linear numerator for fractions with irreducible denominators? For example: (2x^3+5x+1)/((x^2+4)(x^2+x+2)) breaks down to (ax+b/(x^2+4))+(cx+d/(x^2+x+2)). I have been told that since the denominators are irreducible, the numerators will be either linear or constant. Now my question is for something like (2x^3+5x+1/(x^2-4)) you would make it equal (A/(x+2))+(B/(x-2)), why do assume that the numerators are constant? Why couldn't the numerators be linear like the irreducible one?

Another way to think about it is this: all factors are reducible. Specifically, we can break any irreducible quadratic into the factors $(x-(a+bi))$ and $(x-(a-bi))$. Then the partial fraction decomposition will contain $$\frac{C}{x-(a+bi)}+\frac{D}{x-(a-bi)}=\frac{C(x-(a-bi))+D(x-(a+bi))}{(x-a)^2-(bi)^2}=\frac{C'x+D'}{(x-a)^2+b^2}$$ where $C'$ and $D'$ are new constants $$C'=C+D$$ $$D'=-C(a-bi)-D(a+bi)$$ Note that the quadratic $(x-a)^2+b^2$ is irreducible because it has no real roots. Therefore: every pair of complex factors produces a fraction with a linear factor on top and an irreducible quadratic factor on bottom.
