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?

First, suppose that we have a rational function $p(x)/q(x)$.  If $p$ has degree equal to or greater than $q$, we can divide out to get a polynomial quotient plus a fraction $r(x)/q(x)$, with the degree of $r$ less than that of $q$.  Now suppose that we can factorise $q$ as a product of two unequal irreducible polynomials (more than two, I leave as an exercise).  That is, we have
$$\frac{r(x)}{q_1(x)q_2(x)}$$
with $\deg(r)<\deg(q_1q_2)$, and we want partial fractions.
Where this all comes from is the Euclidean algorithm for polynomials.  It's very much like the Euclidean algorithm for integers, and will show (among other things) that if $q_1$ and $q_2$ have no common factor (which is definitely the case if they are irreducible and not equal) then there are polynomials $a$ and $b$ such that
$$r(x)=a(x)q_1(x)+b(x)q_2(x)\ .\tag{$*$}$$
Now there will be infinitely many possibilities for $a(x)$ and $b(x)$, in fact if $a(x)=a_0(x)$ is one specific solution then the general solution for $a(x)$ will be
$$a(x)=a_0(x)+f(x)q_2(x)\ .$$
This means that if we seek the "simplest" value of $a(x)$, the one with smallest degree, we can always make it have degree less than that of $q_2(x)$: whatever $a_0(x)$ we start with, we can take $a(x)$ to be the remainder after dividing by $q_2(x)$.  Then the corresponding $b(x)$ will have degree less than that of $q_1(x)$, for if not then $(*)$ would make $\deg(r)\ge\deg(q_1q_2)$.
Finally, divide $(*)$ by $q_1(x)q_2(x)$ to give
$$\frac{r(x)}{q_1(x)q_2(x)}=\frac{b(x)}{q_1(x)}+\frac{a(x)}{q_2(x)}$$
with $\deg(b)<\deg(q_1)$ and $\deg(a)<\deg(q_2)$ as required.
