<s>[INST] Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $\mathbb Q[x]$ 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.
 [/INST] 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.
</s>