Question about Gauss's Lemma? This is an excerpt from my book:

Suppose that $P(x)$ has integer coefficients and has the zero $x=2/3$ ... By the factor theorem, $P(x)=(x-\dfrac 23)Q(x)$. But what kind of coefficients does $Q(x)$ have? All that we know for sure is that the coefficients must be rational ... We can write $P(x)=(3x-2)S(x)$, where $S(x)= \dfrac {Q(x)}{3}$. We know that $P(x)$ has integer coefficients; can we say the same thing about $S(x)$? Indeed we can; this is Gauss's Lemma:
$$\text{If a polynomial with integer coefficients can be factored into}$$
$$\text{polynomials with rational coefficients, it can also be factored}$$
$$ \text{into primitive polynomials with integer coefficients}$$

My problem is that I don't see how Gauss's Lemma implies that $S(x)$ must have integer coefficients. All that the lemma is saying is that there exist two primitive polynomials $f(x)$ and $g(x)$ such that $P(x)=f(x)g(x)$. I don't know how to deduce that $(3x-2)$ and $S(x)$ are such polynomials.
 A: Gauss lemma can be stated in form that product of primitive polynomials is primitive polynomial. It is not loss of generality to assume that $P$ is primitive. Now, we know that $S(x)$ has rational coefficients, so there exists primitive polynomial with integer coefficients $S_0(x)$ such that $S_0(x) = qS(x)$. Thus, $qP(x) = (3x-2)S_0(x)$, so $qP(x)$ must be primitive which implies that $q = \pm 1$, i.e. $S$ has integer coefficients.

Let me do the general statement. Let $p(x)$ be primitive polynomial with integer coefficients and let $p(x) = f(x)g(x)$ be its factorization over $\mathbb Q$. Then, there exist primitive polynomials with integer coefficients $f_0$ and $g_0$ and rational numbers $q_1$ and $q_2$ such that $q_1f(x) = f_0(x)$, $q_2g(x) = g_0(x)$. Let $q = q_1q_2$. Then we have $qp(x) = f_0(x)g_0(x)$ and by $f_0$ and $g_0$ being primitive, $qp$ is primitive as well. This implies $q=\pm 1$. 
Indeed, let $q = a/b$ with $a$ and $b$ relatively prime and let $\alpha_i$ be coefficients of $p$. First of all, we have that $a\alpha_i/b$ are all integers, so $b$ divides all $\alpha_i$. By assumption of $p$ being primitive, $b = 1$. Finally, we have that $ap$ is primitive, and since $p$ has integer coefficients, $a = \pm 1$. Finally, this gives us factorization $p(x) = \pm f_0(x)g_0(x)$ over $\mathbb Z$.
