Partial fractions and linear vs quadratic factors

Question

I was watching some videos on partial fraction decompistion and I got confused on one of the examples:

Say for example you have $$\frac{x+4}{x^2(x^2 +3)^2}.$$ The partial fraction equation of this is apparently: $$\frac{A}{x} + \frac{B}{x^2} + \frac{Cx+E}{x^2 +3} + \frac{Dx+F}{(x^2 +3)^2}$$

My question is why $A/x+B/x^2$ do not have numerators with an $ax+b$ form, cause $x^2$ is a quadratic not a linear right? Is it because the $x^2$ is in brackets, so you can perceive it as $(x+0)^2$?

Answer 1

Hint: $$\frac{ax+b}{x^2} = \frac{a}{x}+\frac{b}{x^2}.$$ Therefore, $$\frac{A}{x}+\frac{ax+b}{x^2}=\frac{A'}{x}+\frac{b}{x^2}.$$

Answer 2

hint

if $$\frac {x+4}{x^2(x^2+3)^2}=\frac {ax+b}{x}+\frac {cx+d}{x^2}+$$ $$\frac {Ax+B}{x^2+3}+\frac {Cx+D}{(x^2+3)^2} $$

$$=a+\frac {b+c}{x}+\frac {d}{x^2}+... $$

then $x\to +\infty $ gives $a=0$.

Answer 3

You misunderstood one of the assertions in the decomposition theorem: the degree of the numerator has to be less than the degree of the irreducible factor in the denominator.

In the denominator $x^2(x^2+3)$, the irreducible factors are $x$ and $x^2+3$.