Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $\mathbb Q[x]$

Question

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

$x^4-16x^3+20x^2+12$

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.

Answer 1

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

Answer 2

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:

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

$d(1),(4)$:

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

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

Contradiction!

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

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

Answer 3

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 $863 > 1+4+70+500 = 575$.