The question is:


I tried a lot but couldn't solve it so I looked at the solution which is: $$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$ and we can write $$3x^4+2x^3-2x+1=(x^3-x^2+1)+(3x^2-2x)(x^2+x+1)$$ which effectively reduces the integral to very simple ones.

My question is how they deduced the factorization of the denominator. After looking at the solution I think that if we put $x=1,x=\omega$ and $x=\omega^2$ we can deduce this but this was not immediately obvious to me. Is there some sort of hint you can get by looking at the integrand or is it simply a matter of less experience?

Any help would be appreciated.

  • $\begingroup$ a couple tricks you could try that work for these types of problems are $x=1/t$ which is the inversion substitution, or just keep adding zeros in the numerator to get it to equal your denominator’s derivative or your denominator itself. factoring out x’s sometimes works to do that too $\endgroup$
    – C Squared
    Commented Sep 7, 2020 at 12:22

3 Answers 3


Just it's better to see that any polynomial $x^{3k-1}+x^{3n-2}+1$ has a factor $x^2+x+1$ for any naturals $k$ and $n$.

For example, your reasoning with $\omega\neq1$ and $\omega^3=1$ helps to understand it.

In our case we can get this factoring so: $$x^5+x+1=x^5-x^2+x^2+x+1=(x^2+x+1)(x^3-x^2+1).$$

  • 2
    $\begingroup$ More generally, if there are $n$ terms in a polynomial with coefficients all $1$ and taken $\pmod n$ if the exponents cover $0,1,…,n-1$, you can rest assured it is divisible by $(x^n-1)/(x-1)$. May not be easy to recognise always, but it is a simple enough test. +1 $\endgroup$
    – Macavity
    Commented Sep 7, 2020 at 14:02

Hmm. I'd look at it and say "There's no rational root" because $x = \pm 1$ doesn't work. So there's some irrational root, $\alpha$, and two complex-conjugate pairs. So there's no nice obvious linear factor I can write down.

Then I'd say "Maybe there's a quadratic factor." I can assume it's monic, so I'm looking to write $$ x^5 + x + 1 = (x^2 + ax + b) (x^3 + px^2 + qx + r). $$ from which I can expand to get $$ x^5 + x + 1 = x^5+(p + a)x^4 + (q + ap + b) x^3 + (ra + bq)x + br $$ if I've done the algebra right. Equating coefficients I see that \begin{align} 0 &= a + p\\ 0 &= q + ap + b\\ 1 &= ra + bq\\ 1 &= br \end{align} so $ p = -a$, and $r = \frac1b$,and these equations become \begin{align} 0 &= a + -a\\ 0 &= q - a^2 + b\\ 1 &= \frac1b a + bq\\ 1 &= b(1/b) \end{align} which simplify down to \begin{align} q &= a^2 - b\\ b &= a + b^2q\\ \end{align} or \begin{align} q &= a^2 - b\\ 0 &= b^2 q - b + a \end{align}

That last equation is a quadratic in $b$ or $q = 0$. The first choice yields $$ b = \frac{1 \pm \sqrt{1-4aq }}{2q} $$ The second choice yields $b = a, q = 0$, from which we find that $r = 1, p = -1$, which is a nice solution, so we can stop looking at the first case. (Yay!)


try :$$x^5+x+1=(ax^3+bx^2+cx+d)(Ax^2+Bx+C)$$and compare coefficients.!

Again intelligent guessing works the best


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