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?
With a fraction like $ax+b\over cx+d$ it is always possible to remove $a\over c$ copies of the denominator from the numerator, leaving a constant in the numerator:

$${ax+b\over cx+d}={ax+b-ax-\frac ac d\over cx+d}+\frac ac={b-\frac{ad}c\over cx+d}+\frac ac$$

In other words, we have the choice of leaving the fraction in a "linear over linear" form, or changing it to be in a "constant over linear plus constant" form.  Note that a function like $\dfrac a{bx+c}+d$ is easier to graph than one like $\dfrac{ax+b}{cx+d}$ because of the constant offsets compared with the affine transformation involved in the second case.
The answers to partial fraction reductions neglect the following....
(1) a function (numerator) such as ax + c is NOT lineAs previously noted  a polynomial of thaform is affine. Explicitlyan affine function is the composite of a linear function followed by a translation:
ax + b = (x + b) o ax. ax is the linear part, it satisfies the linearity conditis
f(a + b ) = f(a) + f(b) ; f(ka) = kf(a) ( for the experts: a linear function is a vector space homomorphism). A translation does not have those linearity properties, so the composite (affine function) is not linear. If the word "affine" is disliked, "1st. degree polynomial" is acceeptable.
(2) Partial fraction reduction is a consequence of the fact that polynomials  are integral domains, like Z, so possess a Division Algoritm prooperty.
For polnomials f and s, there are two polynomials: q and r such that
f  =  qs + r ; (degree r) &lt; (degree s)....
if (degree f) &lt; (degree s), there is no :q.

and since all polynomials can be factorised  with factors of degree 2 (fundamental theorem of algebra) r can be only of degree o or 1.
The decomposition of a Mobius function given in an earlier answer is treated in C.Sidne'ys Modern Basic Pure Mathematics.