Let $f(x) \in \mathbb Z[x]$ with $c(f)=1$ and $f$ is non-constant. Now suppose $h(x) \in\mathbb Z[x]$ be such that $h(x)=f(x)q(x)$ where $q(x) \in\mathbb Q[x]$. Then I have to show that $q(x) \in\mathbb Z[x]$.
What I have tried by using Gauss lemma there exist $r , s \in\mathbb Q$ such that $f(x) =rf(x)sq(x)$ such that $rf(x) , sq(x) \in \mathbb Z[x]$. But I can't show $r \in\mathbb Z$. I need some help. Thank you.
Assume all coefficients of $q$ are written in reduced terms. Let $d$ be the lcm of the denominators of $q$. Then $dh=fq'$ where $q'=dq\in \mathbb Z[x]$. Since the $c$ function is multiplicative, you get $dc(h)=c(q')$ which shows $d\mid c(q')$. This in particular implies that $d$ divides all coefficients of $q'$, which are the numerators of the coefficients of $q$. So suppose one of the coefficients of $q$ is $a/b$ for some $b>1$ and $(a,b)=1$. Then $b\mid d$, but on the other hand $d\mid a$, so $b\mid a$ which is a contradiction. Therefore $b=1$ and you have the claim.