Show that $f(x)$ is irreducible over $\mathbb{Q}$ $f(x) = x^4 - 10x^2 + 1$.  We can't use Eisenien Criterion since it won't apply to this particular $f(x)$ so i'm not really sure what else to do. Ideas?
 A: By the rational roots theorem, this polynomial has no rational roots.  Therefore, if it is reducible over $\mathbb{Q}$, then it must factor into two quadratics with coefficients in $\mathbb{Q}$.  Such a factorization, if it exists, is not hard to find.  Note that the polynomial is "quadratic" in $x^2$.  
Therefore, you can apply the quadratic formula with the substitution $y = x^2$ and see if the resulting factorization it gives you has rational coefficients.  Applying this method, we get:
$$x^2 = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6}$$
And so the factorization of the polynomial into two quadratics must be:
$$(x^2-(5+2\sqrt{6}))(x^2 + (5+2\sqrt{6}))$$
And so what can we conclude?
A: Since $f$ has no odd powers of $x$, the only possible factorization into quadratics would have the form
$$(x^2+ax+\sigma)(x^2-ax+\sigma)$$
where $\sigma^2=1$ and $2\sigma-a^2=-10$.  But $\sigma^2=1$ implies $\sigma=\pm1$, which leads to $a^2=10\pm2$.  Since neither $12$ nor $8$ is a square, $f$ does not have any quadratic factors over the rationals.
The absence of odd powers and the fact that $f(0)\not=0$ now implies that $f$ has no linear factors either:  If $x-r$ were a factor, then $x+r$ would also be a factor, in which case $x^2-r^2$ would be a quadratic factor.  But we just proved that $f$ has no quadratic factors.
