Monic Factors in $\mathbb Q[x]$ of a Monic $f \in \mathbb Z[x]$ are also in $\mathbb Z[x]$ 
The claim is that if $f \in \mathbb Z[x]$ is monic and has a factorization, say $f = pq$ with $p,q \in \mathbb Q[x]$ such that $p, q$ are also monic, then $p, q$ are in $\mathbb Z[x]$.

The text cited Gauss's lemma but I am not seeing how.  Suppose that $R$ is a unique factorization domain, and $F$ its field of quotients; Gauss's lemma says that, if $f \in R[x]$ is primitive, reducibility over $F$ is equivalent to reducibility over $R$.  However, the statement does not say that the factorizations are the same.
The premise that $f$ is monic suggest that the content $c(f) = 1$.  Thus, I attempted a proof by contradiction.  What happens if the factors are not in $\mathbb Z[x]$, but here is the problem.  Since each rational number has infinitely many factors, how do I know what $c(p), c(q)$ equal to?
 A: Assume that a polynomial $f\in \mathbf{Z}[x]$ that does not factor in ${\bf{Z}}[x]$ has a proper factorization $f=gh$ in ${\bf{Q}}[x]$. Then multiplying both sides of $f=gh$ by the product of the least common multiples of the denominators of the coefficients of $g$ and $h$ for example gives us a positive integer $n$ such that $nf=g_1h_1$ where $g_1$ and $h_1$ are both in ${\bf{Z}}[x]$. Let $n_0$ be the smallest positive integer such that ${n_0}f=g_1h_1$, where $g_1$ and $h_1$ are both in ${\bf{Z}}[x].$ If $n_0 =1$, there is nothing to prove. So assume that $n_0 > 1$ and let $p$ be a prime divisor of $n_0$. We contend that either $p$ divides ${\it every}$ coefficient of $g_1$, or that $p$ divides ${\it every}$ coefficient of $h_1$. Assume that this is not so, and for $g_1(x)={{\sum}^{m}_{i=0}}{a_i}x^i$, and $h_1(x)={{\sum}^{n}_{i=0}}{b_i}x^i,$ let $k$ be the smallest positive integer for which $p$ does not divide $a_k$, and $l$ be the smallest positive integer for which $p$ does not divide $b_l$. Then since $p$ divides $n_0f$, if must divide the coefficient $${{\sum}^{k}_{i=0}}{a_i}{b_{(k+l)-i}}.$$ This forces us to conclude that $p$ divides ${a_k}{b_l}$ which is impossible since $p$ is a prime and since $p$ does nor divide both $a_k$ and $b_l$. So, we have factorization $({n_0}/p)f={g'_1}{h'_1}$ where $g'_1$ and $h'_1$ are both in ${\bf{Z}}[x]$. This in turn contradicts our choice of $n_0$ as being the smallest positive integer for which a factorization as above is possible. Thus the factorization $f=gh$ was a factorization in ${\bf{Z}}[x]$ to begin with, and so a factorization of an irreducible polynomial $f(x)\in{{\bf{Z}}[x]}$ is not possible in ${\bf{Q}}[x].$ This argument applies to monic polynomials as well.
