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?
 A: 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.
A: 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.
A: 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.
A: The reason that we always make a partial fraction's numerator at least $1$ degree less than that of the denominator is that, if the numerator's degree is greater than or equal to that of the denominator, the highest degree can be factored out as a non-fraction term.
Taking your $\displaystyle \frac{2x^3+5x+1}{x^2-4}$ for example, we get
$$\frac{2x^3+5x+1}{x^2-4} = \frac{2x^3 - 8x + 13x + 1}{x^2-4} = \frac{2x (x^2 - 4) + 13x + 1}{x^2-4} = 2x + \frac{13x + 1}{x^2-4}$$
$$= 2x + \frac{A(x+2)+B(x-2)}{(x-2)(x+2)} = 2x + \frac{A}{x-2} + \frac{B}{x+2}$$
So $13x+1 = A(x+2) + B(x-2)$. I leave the math now for you to solve.
A: 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) < (degree s)....
if (degree f) < (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. 
