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? )


  • $\begingroup$ Welcome to Math Stack Exchange. Please use MathJax. Consider factoring $9x^2-1$ $\endgroup$ – J. W. Tanner Mar 6 '19 at 23:19
  • 2
    $\begingroup$ 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. $\endgroup$ – 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.


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