Trying to prove that the following polynomial is irreducible in $\mathbb Q[x]$:


What I have tried:

1.) Eisenstein's Criterion, but there exists no suitable prime.

2.) reducing to modulo 2, 3, 5 ,7, 11, but by my calculations, reduction to mod 2, mod 3, mod 5, mod 7, yields a reducible polynomial. Mod 11 seems like it could potentially work, but I can't believe that would be the correct approach, given the sheer number of potential quadratic factors one would have to check.

3.) This polynomial fails the rational roots test, so I know that the only possible factors would involve second degree polynomials. Guided by some of the previous posts on related questions, I have attempted to work out some type of contradiction by assuming the polynomial can be factored like $(x^2+ax+b)(x^2+cx+d)$, but I haven't had much luck with this approach.

I imagine I'm staring at something obvious but can not see it. Any help would be appreciated.

  • 1
    $\begingroup$ Do you know Gauss' Lemma? $\endgroup$
    – Sarastro
    May 17 '15 at 15:18
  • $\begingroup$ @user8268, doesn't the Newton polygon have rise=2 and run=4, so has (2,1) in its interior? $\endgroup$ May 17 '15 at 15:25
  • $\begingroup$ @paulgarrett and indeed I cannot draw :) $\endgroup$
    – user8268
    May 17 '15 at 15:28
  • $\begingroup$ @user8268, :) But, yes, 2-adically there are either two irreducible quadratic factors, or an irred quartic. $\endgroup$ May 17 '15 at 15:31

Note that any factorization will be of the form: $$x^4 -16x^3+20x^2+12 = (x^2+2ax+2b)(x^2+2cx+2d)$$ Here $a,b,c,d$ are integers. We have used Gauss' lemma to note that if $f(x)$ is reducible over $\Bbb Q$ then it is reducible over $\Bbb Z$. The factors of two come from the fact that after reducing mod $2$, our factorization must turn into a factorization of $x^4$.

Expanding our expression gives us the following equations to be satisfied. $$\begin{align*} 2a + 2c &= -16\\ 2b+2d + 4ac &= 20\\ 4bc+4ad &= 0\\ 4bd &= 12 \end{align*}$$ The last of these implies that the pair $\{b,d\}$ is either $\{1,3\}$ or $\{-1, -3\}$. From this setup it is elementary to check that no choice of integers will work

  • $\begingroup$ So we're assuming that this is reducible over Q, and then by Gauss, it would be reducible over Z, and showing that since it is not reducible over Z, it can not be reducible over Q? With the restricted choices for b,d then a and c can't be integers, and so it can not be factored as such. The factors of 2 are included so when expanded we have $x^2+2(c+a)x^3+2(d+2ac+b)x^2+2(2ad+2bc)x+2(2bd)$, which ensures that when we reduce this mod 2, every term drops out and we are left with $x^4$, as is the case when we reduce the original polynomial mod 2. Is this the correct interpretation? $\endgroup$
    – mrmingus
    May 17 '15 at 20:04
  • $\begingroup$ @mrmingus the argument for the factors of 2 is a bit more involved. Let $f(x) = g(x)h(x)$. Then $\overline{f}(x) = \overline{g}(x)\overline{h}(x)$, where $\overline{f}, \overline{g}, \overline{h}$ are the corresponding polynomials with coefficients reduced mod $2$. However, we already know that $\overline{f} = x^4$ and $\Bbb F_2[x]$ is a UFD, so that we must have $\overline{g} = x^2, \overline{h} = x^2$ if they are both quadratic. $\endgroup$
    – Rolf Hoyer
    May 17 '15 at 20:17
  • $\begingroup$ OK. Need to think that over for a bit. Thanks for the response. $\endgroup$
    – mrmingus
    May 17 '15 at 22:18

I am here to prove by Contradiction.

Assume $x^4-16x^3+20x^2+12$ can be factorized, so we have:

\begin{align} x^4-16x^3+20x^2+12&=(x^2+ax+b)(x^2+cx+d)\\&=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd \end{align}

By comparing similar terms, we have:

$$ad+bc=0------(1)$$ $$ac+b+d=20----(2)$$ $$a+c=-16-----(3)$$ $$bd=12-------(4)$$


$$ad^2+bdc=0$$ $$d=\pm{\sqrt{\frac{-12c}a}}$$

For $d\in{\Bbb{Q}}, \exists{k\in{\Bbb{Q}}\text{\0}}$, $\frac{c}a=-3k^2----(5)$,

so $d=\pm{6k}-----(6)$

Sub (5),(6) into $(1)/a$,

$$\pm{6k}+b(-3k^2)=0$$ $$b=\pm\frac2k$$

Sub (5) into (3), $$a-3ak^2=-16$$ $$a=\frac{16}{3k^2-1}$$ $$c=-3ak^2=\frac{-48k^2}{3k^2-1}$$

Sub $a,b,c,d$ to original expression,

\begin{align} (x^2+ax+b)(x^2+cx+d)&=(x^2+\frac{16}{3k^2-1}x\pm\frac2k)(x^2+\frac{-48k^2}{3k^2-1}x\pm{6k})\\&=((3k^2-1)x^2+16x\pm{\frac{2(3k^2-1)}{k}})((3k^2-1)x^2-48k^2\pm{6k(3k^2-1)}) \end{align}

So $$12(3k^2-1)^2=12$$ $$3k^2=\pm{1}+1$$ $$k^2=\frac23\text{ or } 0 \text{ (rej.) }$$ $$k=\pm{\sqrt{\frac23}}\notin{\Bbb{Q}}$$

Recall For $d\in{\Bbb{Q}\text{\0}}, \exists{k\in{\Bbb{Q}}}$


$x^4-16x^3+20x^2+12$ cannot be factorized into expression in which $k\in{\Bbb{Q}}$.

Therefore the irreducibility of $x^4-16x^3+20x^2+12$ is proven.

  • $\begingroup$ Brute force + Contradiction. $\endgroup$
    – user26857
    May 17 '15 at 17:15
  • $\begingroup$ This is suggested in pt (3) by the OP. Just for reference if he/she wants it. $\endgroup$ May 17 '15 at 17:16

Here are in my opinion interesting prime related criteria, although it might take some effort to find suitable primes.

First approach, we can use criterion of Murty (as used in this other answer of mine for example Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible), for reverted polynomial $g(x) =1/x^4f(1/x) = 12x^4+20x^2-16x+1$, since $f(4)=3329$ is a prime.

Or we can also use criterion given by Osada in Prasolov's book Polynomials (as shown in this answer for example Proving irreducibility of $x^6-72$), this time for $f(x+5)=x^4+4x^3-70x^2-500x-863$, since $863$ is a prime and $863 > 1+4+70+500 = 575$.


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