# Irreducibility of Non-monic Quartic Polynomials in Q[x]

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.

• $f(x)$ has an easy linear factor. – almagest May 3 '16 at 19:42
• Try $x=-3$ for $f(x)$. – Dietrich Burde May 3 '16 at 19:46
• Yes, but how would one determine that f(x) is reducible other than plugging in random numbers? – Jason Smith May 3 '16 at 19:49

## 2 Answers

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.

• How did you know -3 is a root? – Jason Smith May 3 '16 at 19:58
• 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). – Gro-Tsen May 3 '16 at 20:01
• @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 $\pm 3$.) – Gro-Tsen May 3 '16 at 20:04

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.

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