My question comes from this specific problem from my homework:
$$\frac{x}{81x^4 - 1}$$
Initially, I factored the denominator out to $(9x^2+1)(9x^2-1)$ and used this to find the $A,B,C,D$ to decompose the fraction, and everything seemed to be working fine but when I entered the final answer into the homework website it said it was wrong. The website had a video showing how to get the answer and they factored the denom. out into the $3$ factors instead of $2$.
This is probably the only thing that annoys me about partial fractions. Sometimes I can't tell if I should factor the denom. into $2$, $3$, or $4$ factors. If there is some sign I am missing here please do tell!
(Also, I do know that usually, you should factor it so there are as many factors as there are terms in the numerator. Or is that wrong? )
Thanks,
The number of factors is not important. What is important is to decompose the denominator into irreducible factors. And, if you are working over $\mathbb R$ or over $\mathbb Q$, then, although $9x^2+1$ is irreducible, $9x^2-1$ is not, since it is equal to $(3x-1)(3x+1)$.
And if you were working over $\mathbb C$, then you would have to write $9x^2+1$ as $(3x-i)(3x+i)$ too.