Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ 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?
The associated quadratic polynomial $t^2-4t+16$ has negative discriminant, so there's no real root. Then the polynomial can be factorized over the reals as a product of degree two polynomial. You get them by a process similar to completing the square:
\begin{align}
x^4-4x^2+16
&amp;=x^4+8x^2+16-12x^2\\
&amp;=(x^2+4)^2-(\sqrt{12}\,x)^2\\
&amp;=(x^2-\sqrt{12}\,x+4)(x^2+\sqrt{12}\,x+4)
\end{align}
These two polynomials have negative discriminant (no need to verify it) and so they're irreducible in $\mathbb{R}[x]$. If the given polynomial were reducible over the rationals, the two factorizations in $\mathbb{Q}[x]$ and $\mathbb{R}[x]$ would coincide.

Therefore the given polynomial is irreducible over the rationals.



What's the general rule? Suppose you have $x^4+px^2+q$, with $p,q$ integers and $p^2-4q&lt;0$ (so $q&gt;0$). Write $q=r^2$, with $r&gt;0$ (it need not be integer), and
$$
x^4+px^2+q=x^4+2rx^2+r^2-(2r-p)x^2
$$
Note that $2r-p&gt;0$: it's obvious if $p&lt;0$; if $p\ge0$ it's the same as $4q&gt;p^2$, which is true by hypothesis. Then
$$
x^4+px^2+q=(x^2-\sqrt{2r-p}\,x+r)(x^2+\sqrt{2r-p}\,x+r)
$$
is the decomposition of the polynomial in $\mathbb{R}[x]$. It is in $\mathbb{Q}[x]$ if and only if $\sqrt{q}$ and $\sqrt{2\sqrt{q}-p}$ are integers.

For example, $q=4$ and $p=0$ is a case. For $q=16$ we need $8-p$ to be a square, so $q=16$ and $p=4$ is another case.
f(x) = 0 leads to $x^4$ +16 = 4$x^2$ then for |x| = 0,1, 2 and for |x| ≥ 3 the equality is impossible.Then x can't be integer. On the other hand we have 
($x^2$ - 2)^2 = -12 which gives $x^2$ = 2(1 +sqrt(-3)) and x can't be rational. What remains is to prove that f has not two quadratic factors of Z[x]  wich can be proved with undetermined coefficients. I stop here.