If $g(x),\; h(x)$ are irreducible over the rationals, then the Galois group of $f(x) = g(x)h(x)$ permutes the roots of $g$ and the roots of $h$ amongst themselves, separately. In other words no root of $g$ can be sent to a root of $h$ or vice versa under action of the Galois group of $f$ over the rationals?
Is this true? It seems true but I'm wary that there might be some special case that I have overlooked.