Why the linear numerator for fractions with irreducible denominators and constant numerators for reducible denominators?

Question

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??

Answer 1

First you divide the denominator into the numerator, getting a remainder. So, for your example, $$\frac{2x^3+5x+1}{x^2-4}=2x+\frac{13x+1}{x^2-4}$$ and expand the last fraction in partial fractions. As the denominator now has a higher degree than the numerator, the limit as $x \to \infty$ is $0$, so you don't want a constant term. In fact, $$\frac{13x+1}{x^2-4}=\frac {25/4}{x+2}+\frac{27/4}{x-2}$$ and you don't want or need a linear term in the numerator.

Answer 2

That is actually incorrect... you need to do polynomial long division first.

Answer 3

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 $$\frac{2x^3+5x+1}{(x+2)(x-2)}=\frac{ax+b}{x+2}+\frac{cx+d}{x-2}=\frac{ax+b-a(x+2)}{x+2}+a+\frac{cx+d-c(x-2)}{x-2}+c=\frac{ax+b-ax-2a}{x+2}+\frac{cx+d-cx+2c}{x-2}+a+c=\frac{b-2a}{x+2}+\frac{2c+d}{x-2}+a+c$$