I know from a lemma in Herstein that if a primitive polynomial in $R[x]$ is irreducible in its field of quotients $F[x]$, then it is irreducible in $R[x]$.
But, if some non-primitive polynomial in $\mathbb Z[x]$ is irreducible over $\mathbb Q$, does this imply the non-primitive polynomial is irreducible over $\mathbb Z$?
By definition an element $f$ of a ring $R$ is irreducible if whenever $f=uv$ for $u,v\in R$, either $u$ or $v$ is a unit.
By definition if $f$ is nonprimitive in $\Bbb{Z}[x]$, $f=ng(x)$ where $n$ is the gcd of it's coefficients, which is not $\pm 1$, and hence not a unit, and $g(x)$ is a primitive polynomial in $\Bbb{Z}[x]$ of positive degree and hence not a unit. Thus regardless of whether or not $f$ is irreducible in $\Bbb{Q}[x]$, it is reducible in $\Bbb{Z}[x]$.
Note that the constant becomes a unit in $\Bbb{Q}[x]$, so it doesn't matter there, but it does matter in $\Bbb{Z}[x]$.