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?

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.
