Why the linear numerator for fractions with irreducible denominators and constant numerators for reducible denominators? For example:
$\Large{\frac{2x^3+5x+1}{(x^2+4)(x^2+x+2)}}$ breaks down to $\Large{\frac{ax+b}{x^2+4}+\frac{cx+d}{x^2+x+2}}$
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 $\Large{\frac{2x^3+5x+1}{x^2-4}}$ you would make it equal $\Large{\frac{A}{x+2}+\frac{B}{x-2}}$, why do we assume that the numerators are constant? Why couldn't they be linear??

Because remainder of division is always smaller then the degree of the polynomial so we assume it's exactly 1 smaller,and find if the other terms are zero
$$\frac{2x^3+5x+1}{(x+2)(x-2)}=\frac{ax+b}{x+2}+\frac{cx+d}{x-2}=\frac{ax+b-a(x+2)}{x+2}+a+\frac{cx+d-c(x-2)}{x-2}+c=\frac{ax+b-ax-2a}{x+2}+\frac{cx+d-cx+2c}{x-2}+a+c=\frac{b-2a}{x+2}+\frac{2c+d}{x-2}+a+c$$
