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Which way I can determine whether the polynomial $x^5 + 5x^2 +1$ is irreducible over $\mathbb Q$ or not?

Mod $p$ Irreducibility Test and Eisenstein's criterion not applicable here.

Which way I should proceed?

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  • $\begingroup$ You mean you cant use the mod P irreducibility or that you tried without success? $\endgroup$ – AHandsomeAlien Jun 29 '17 at 4:29
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    $\begingroup$ Reduce modulo $2$. Show that your new polymomial has no roots in $\Bbb Z_2$ and show that your reduced polynomial can't be written as $q(x)p(x)$ where $q$ has degree $2$ and $p$ has degree $3$. $\endgroup$ – Rocket Man Jun 29 '17 at 4:31
  • $\begingroup$ No. Actually I tried. But then after reducing by mod5 polynomial becomes x^5 + 1 . Then I dont know how should I check if x^5 +1 is irreducible in Z5 or not. $\endgroup$ – user458361 Jun 29 '17 at 4:32
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    $\begingroup$ I second AJ Stas's advice. Reduce modulo 2. The polynomial $x^5+x^2+1$ has no zeros in $\Bbb{Z}_2$, and it is not divisible by the unique (!) quadratic irreducible either. $\endgroup$ – Jyrki Lahtonen Jun 29 '17 at 4:41
  • $\begingroup$ By the way, $x^5 + 1$ is not irreducible mod $5$ because $-1$ is a root. $\endgroup$ – cat Jun 29 '17 at 7:29
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Try $x=y-1$. It should help by the Eisenstein's criterion.

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  • $\begingroup$ Yeah it worked!! Thanks Dear!! $\endgroup$ – user458361 Jun 29 '17 at 4:47
  • $\begingroup$ @user458361 You are welcome! $\endgroup$ – Michael Rozenberg Jun 29 '17 at 4:59
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$7^5+5\cdot 7^2+1 = 17053$ which is a prime number. Thus the polynomial is irreducible by Cohn's irreducibility criterion.

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