Let $f,g,h \in F[x]$, with $f(x)$ and $g(x)$ relatively prime. If $f(x)$ divides $h(x)$ and $g(x)$ divides $h(x)$ prove that $f(x)g(x)$ divides $h(x)$.
My thoughts: there are certain properties that prove that multiplied together these will divide $h(x)$.
If both $f(x)$ and $g(x)$ divide $h(x)$, $h(x)$ includes all of both of their factors. The factorization of $f(x)g(x)$ will obviously include all the factors of both. If $f(x)$ and $g(x)$ were not relatively prime, the factors they shared would be compounded in the factors of $f(x)g(x)$ and there is no guarantee that the factorization of $h(x)$ would include this many of these factors. However since there is no overlap between the factors of $f(x)$ and $g(x)$, the factors of $f(x)g(x)$ must simply be the union of the factors of each, which $h(x)$ must include.