I am trying to factor the polynomial $2X^{10}+4X^5 +3$ into irreducible factors in $(\mathbb{Z} / 5\mathbb{Z})[X].$

I have already determined that the polynomial has no linear factor because $p(0) \neq 0,$ $p(1) \neq 0,$ $p(2) \neq 0,$ $p(3) \neq 0$ and $p(4) \neq 0.$

However, I am not sure how to continue.

Any help would be appreciated.

  • 4
    $\begingroup$ Don't forget: $\pmod 5$ we have $(a+b)^5=a^5+b^5$. That makes this problem very easy. $\endgroup$
    – lulu
    Jul 1 '18 at 22:53
  • 1
    $\begingroup$ @lulu I think your comment serves as great answer. Please post it as an answer; I'll upvote it. $\endgroup$
    – amWhy
    Jul 1 '18 at 22:57

Use the fact that we have $(a+b)^5\equiv a^5+b^5\pmod 5$.

Specifically: $$2x^{10}+4x^5+3\equiv 2\,(x^{10}+2x^5+4)\equiv 2\,(x^2+2x+4)^5\pmod 5$$

It remains to show that $x^2+2x+4$ is irreducible $\pmod 5$ but that follows from what you have already done (or by direct computation).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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