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.