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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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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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).

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