Proposition: The polynomial $(x^4+1)$ is reducible in the ring $(\mathbb Z/p \mathbb Z)[x]$ for every prime $p$.

Proof:

If $p=2$ then $x^4+1 = (x+1)^4$ is clearly reducible. Otherwise, let $p$ be an odd prime. Then, $p^2\equiv 1 (\mod 8)$, so

$(x^4+1) | (x^8-1) | (x^{{p^2}-1}-1) | (x^{p^2}-x)$

Hence the roots of the given polynomial lie in the field $\mathbb F_{p^2}$. Had the quartic been irreducible, the roots would have been found in an extension of degree 4 over $\mathbb F_p$.

Proposition: The polynomial $(x^2-2)(x^2-3)(x^2-6)$ has a root modulo every prime $p$, that is, in $(\mathbb Z/p \mathbb Z)[x]$.

Proof:

If $\displaystyle\left(\frac{2}{p}\right) = \left(\frac{3}{p}\right) = -1$, then by multiplicativity of the Legendre symbol, $\displaystyle\left(\frac{6}{p}\right) = 1$.