Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I prove that $x^5+6x^3+x^2+3x+2$ is irreducible in $\mathbb{Q}[x]$?

I tried with Eisenstein (also making the substitution $x\mapsto x-1$ and $x\mapsto x+1$ to see if I obtain an Eisenstein polynomial) but nothing. If we go modulo 2 we obtain a reducible polynomial, and modulo 3 we obtain $x^5+x^2+2$ and I don't know how to prove that it is irreducible.

Any help?

share|cite|improve this question
There is always brute force. Modulo 3 there are only so many polynomials it could reduce to a product of, so you might check them all. – Jon Beardsley Dec 30 '11 at 3:18
It's irreducible in $\mathbb F_{41}$. – user16697 Dec 30 '11 at 3:18
@JBeardz - Is the brute force the only way to solve this problem? – Victor Dec 30 '11 at 3:19
mod 3, isn't $x^5 + x^2 + 2 \equiv 2(x^2+1)$? $x^2+1$ is irreducible over $\mathbb{Q}$... the equivalence is by Fermat – mathmath8128 Dec 30 '11 at 3:21
@Victor No not at all. – Jon Beardsley Dec 30 '11 at 3:27
up vote 10 down vote accepted

By the rational root theorem the only possible rational roots are $\pm 1, \pm 2$, and by inspection none of these are roots. If the polynomial is reducible, it therefore factors into the product of a quadratic and cubic factor (over $\mathbb{Z}$ by Gauss's lemma).

$\bmod 2$ the polynomial factors as $x(x^4 + x + 1)$. The latter factor has no root $\bmod 2$, so if it is reducible it is the product of two irreducible quadratics. But the only irreducible quadratic $\bmod 2$ is $x^2 + x + 1$, and $(x^2 + x + 1)^2 = x^4 + x^2 + 1$. Hence $x^4 + x + 1$ is irreducible $\bmod 2$.

But if the polynomial factored as the product of a qudaratic and cubic factor over $\mathbb{Z}$, it would only have at most cubic irreducible factors $\bmod 2$; contradiction. Hence the polynomial is irreducible.

share|cite|improve this answer
Scooped me by a minute...*precisely* the same answer! – Bill Dubuque Dec 30 '11 at 3:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.