Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible 
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.

I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:

$f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$
For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2

Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method...  so have any idea how we can prove that $f$ is irreducible?

$$x^4-4x^2+16=(x^2-(2+\sqrt{-12}))(x^2-(2-\sqrt{-12}))$$
No rational roots and no factorization into quadratics over the rationals. The polynomial is irreducible over the rationals
edit for those who commented that this is not enough. I factorized over $\mathbb{C}[X]$ and thus proved that there are no rational solution i.e no degree $1$ factors. The only factorization possible is therefore into two quadratics. $\mathbb{C}[X]$ is a UFD and therefore we have 
$$x^4-4x^2+16=(x-\sqrt{2+\sqrt{-12}})(x+\sqrt{2+\sqrt{-12}})(x-\sqrt{2-\sqrt{-12}})(x+\sqrt{2-\sqrt{-12}})$$
And this is unique. So combining the degree 1 factors in pairs is the only way to factorize in quadratics and there are three different ways to combine and none is rational
