I have one question:
Let's say that in $\Bbb Q[x]$ the element $(x^2 - 2)$ by the mod $p$ test, for $p= 2$ reduces to $x^2$ which is reducible over $\Bbb Q[x]$. But we know from factorization of $(x^2 - 2)$ can be expressed as $(x - \sqrt2)(x +\sqrt2)$ which has irrational roots and hence $x^2 - 2$ is irreducible.
Does the mod $p$ test fail here, or am I missing something?
The "mod $p$ test" gives a necessary but not sufficient condition for a polynomial to be reducible (or, contrapositively, a sufficient but not necessary condition for a polynomial to be irreducible). That is, if a monic polynomial $f(x)\in\mathbb{Z}[x]$ is reducible, then the image of $f(x)$ in $\mathbb{Z}/p[x]$ is also reducible (since you can just take the image of each of the factors). But, as your example shows, the converse is not necessarily true: it might be that the image of $f(x)$ in $\mathbb{Z}/p[x]$ is reducible even if $f(x)$ is not reducible.