Proving that $x^4 - 10x^2 + 1$ is not irreducible over $\mathbb{Z}_p$ for any prime $p$.

We have $$x^4-10x^2+1 =(x^2-1)^2-8x^2 = (x^2-1-2ax)(x^2-1+2ax)$$ if $2=a^2$, or $$x^4-10x^2+1 =(x^2+1)^2-12x^2 =(x^2+1-2bx)(x^2+1+2bx)$$ if $3=b^2$. Thus the only cases that remain are those where neither $2$ nor $3$ is a square - but then $6$ is a square and with $6=c^2$, we find $$x^4-10x^2+1 =(x^2-5)^2-24=(x^2-5-2c)(x^2-5+2c).$$

(Note that we do not even need quadratic reciprocity or similar "advanced" stuff - all we need is that the product of non-squares is square, which follows from the fact that $\Bbb F_p^\times$ is cyclic).

Another simple explanation comes from the fact that the zeros of $m(x)=x^4-10x^2+1$ are $$ x=\pm\sqrt2\pm\sqrt3 $$ with all four sign combinations. So if $p$ is a prime, then you get the splitting field of $m(x)$ over $K=\Bbb{F}_p$ by adjoining $\sqrt2$ and $\sqrt3$. Because up to isomorphism the field $K$ has only a single quadratic extension, namely $L=\Bbb{F}_{p^2}$, we immediately see that $m(x)$ splits into linear factors over $L$. This is because $\sqrt2$ and $\sqrt3$ are both elements of $L$. Consequently $m(x)$ splits into quadratics at worst over $K$.

From this way of looking at it it is obvious how to generalize this. Any biquadratic polynomial with zeros of the form $\pm\sqrt{d_1}\pm\sqrt{d_2}$ for some integers $d_1,d_2$ will split into quadratic (or linear) factors modulo $p$ for all primes $p$. From basic field theory we see that it will be irreducible over $\Bbb{Q}$ whenever the field $F=\Bbb{Q}(\sqrt{d_1},\sqrt{d_2})$ is a degree four extension over $\Bbb{Q}$.

See Qiaochu's answer here for more Galois theory.