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Decide whether the following polynomials are reducible or not in the given ring.

$f(x)=5x^4+22x^3+35x^2+47x+15$ in Q$[x]$

$g(x)=2x^4-4x^3-12x^2+14x-2$ in Q$[x]$

I am not sure what method to use to determine irreducibility here. I know, for quadratic and cubic polynomials, the polynomial is irreducible if it has no roots in Q$[x]$. But of course these are quartic polynomials. I cannot use the Eisenstein Criterion because no prime number divides the coefficients of $f(x)$ and the only prime number, 2, that divides the coefficients of $g(x)$ also divides the leading coefficient, which is not acceptable when using the Criterion. I would also want to convert the coefficients of each to modulo 2 so that I could show they are irreducible in F$_2$[x], but this only works for monic functions. Any idea of what I can do?

Thanks.

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  • $\begingroup$ $f(x)$ has an easy linear factor. $\endgroup$ – almagest May 3 '16 at 19:42
  • $\begingroup$ Try $x=-3$ for $f(x)$. $\endgroup$ – Dietrich Burde May 3 '16 at 19:46
  • $\begingroup$ Yes, but how would one determine that f(x) is reducible other than plugging in random numbers? $\endgroup$ – Jason Smith May 3 '16 at 19:49
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Your $f$ is not irreducible because $-3$ is a root.

As for $g$, first you should divide it by $2$ because it is not primitive (in $\mathbb{Z}[x]$ it is not irreducible, but of course for $\mathbb{Q}[x]$ this is irrelevant); after dividing by $2$, if you reduce mod $2$ you get $x^4 + x + 1$ which is irreducible in $\mathbb{F}_2[x]$ (as is easily checked by trying all possible factors of degrees $1$ and $2$), so $g$ is irreducible.

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  • $\begingroup$ How did you know -3 is a root? $\endgroup$ – Jason Smith May 3 '16 at 19:58
  • $\begingroup$ Frankly? Because I fed it into Sage and Sage told me that. But it makes sense to try the small integers to see if they are roots. Of course, if it had been $37$ the root, things would have been different (but there are bounds on how large the factors can be: search for "Mignotte's bound"; and, of course, the whole process of factorization of integer polynomials is algorithmic so it can, in theory, be performed by hand). $\endgroup$ – Gro-Tsen May 3 '16 at 20:01
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    $\begingroup$ @JasonSmith (And, of course, the integers you have to try aren't too many, since they have to divide $15/5=3$, so you just had to try $\pm1$ and $\pm 3$.) $\endgroup$ – Gro-Tsen May 3 '16 at 20:04
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The polynomial $g(x)$ can be written as $2h(x)$ with a monic polynomial $h(x)$ which is irreducible. Since it is monic you can try the different methods. $f(x)$ is obviously reducible since $(x+3)$ is a linear factor.

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  • $\begingroup$ Ok, I understand that g(x) is irreducible, but how did you know that (x+3) is a linear factor of f(x)? $\endgroup$ – Jason Smith May 3 '16 at 19:54
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    $\begingroup$ @JasonSmith rational roots theorem $\endgroup$ – Will Jagy May 3 '16 at 19:59

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