How to check whether $f(X)=X^4-6X^2+1$ is irreducible or not in $\mathbb{Q}[X]$?
I have tried to apply Eisenstein's Criterion to a linear substitution, to reduce the $f(X)$ modulo a prime and to write it as $(X^2+aX+c)(X^2+bX+d)$, but all without success.
I know that $f(X)$ has four zeros, namely $\pm\sqrt{\frac{6\pm 4\sqrt{2}}{2}}$, these are all not in $\mathbb{Q}$, but I don't think we can conclude anything from this.
Any hints on how to approach this problem? Or on what method to use when I get stuck like this?
An idea that just crossed my mind was to reduce modulo a prime and then try to write it as something of the form $(X^2+aX+c)(X^2+bX+d)$, which I can see doesn't work for $p=2$, as (X^4+X^2+1)=(X^2+x+1)^2 in $\mathbb{F}_2[X]$, I haven't tried to work any others out yet.
The monic irreducible factors in $\mathbf Q[x]$ of a monic polynomial in $\mathbf Z[x]$ are in $\mathbf Z[x]$ (Gauss's lemma). There are no roots in $\mathbf Z$, so if the polynomial were reducible in $\mathbf Q[x]$ then it would equal $$ (x^2 + ax + c)(x^2 + bx + d) $$ in $\mathbf Z[x]$, so $cd = 1$. Thus $c = d = 1$ or $c = d = -1$. Try each of those cases to solve for $a$ and $b$ as integers. If it works out in one case then you have shown the polynomial is reducible. If it doesn't work in either case then you have shown the polynomial is irreducible.