# Is there a way to tell how to factor the denom. when doing a partial fraction?

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,

• Welcome to Math Stack Exchange. Please use MathJax. Consider factoring $9x^2-1$ – J. W. Tanner Mar 6 '19 at 23:19
• Web-based assignment software is often useless, or worse: you might well have got a correct answer, just not the correct answer that it was expecting, and unlike a human, there's no leeway for alternative but equally correct answers. Presumably, the expected answer was the one in which the denominator had been factorised as far as it could go: $81x^4 - 1 = (9x^2+1)(3x-1)(3x+1)$ (notice you can't factorise any of these brackets further). But that doesn't mean your answer was wrong necessarily. It depends on the question and the context. – Billy Mar 6 '19 at 23:20

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.