Reducibility over $Q$ implies reducibility over $Z$

$Q$ : ring of rational number (which is field)
$Z$ : ring of integer (which is integral domain, not a field)
content of polynomial : greatest common divisor of the coefficients of polynomial (must be nonzero polynomial)

Suppose $f(x) \in Z[x]$ and $f(x) = g(x)h(x)$, where $g(x),\ h(x) \in Q[x]$ and $1 \leq deg\ g(x),\ h(x) < deg\ f(x)$. Also, we can assume that $f(x)$ is primitive polynomial. Since that dividing both sides with the content of $f(x)$, say $c_{f(x)}$, derives 

$$\frac{f(x)}{c_{f(x)}} = \frac{g(x)h(x)}{c_f(x)}$$

Which makes polynomial in left side primitive, but does not affect degree of $g(x)$ nor $h(x)$. Hence, $f(x)$ can be primitive in any situation.

Let $a$ be the LCM(Least Common Divisor) of the denominators of the coefficients of $g(x)$, and $b$ be the LCM of the denominators of the coefficients of $h(x)$.
Then, $abf(x) = ag(x) \cdot bh(x)\ (ag(x),\ bh(x) \in Z[x])$.

Let $c_{1}$ be the content of $ag(x)$, and $c_{2}$ be the content of $bh(x)$.
Then, we can write $ag(x) = c_{1}g^{'}(x)$ and $bh(x) = c_{2}h^{'}(x)$.

Which implies that $g^{'}(x)$ and $h^{'}(x)$ are primitive polynomial,
and $abf(x) = c_{1}c_{2}g^{'}(x)h^{'}(x)$.

Since $f(x)$ is primitive polynomial, the content of $abf(x)$ is $ab$.
Since, moreover, by Gauss's Lemma (in algebra), product of primitive polynomial is primitive polynomial. Hence, the content of $c_{1}c_{2}g^{'}(x)h^{'}(x)$ is $c_{1}c_{2}$.

Thus, $ab = c_{1}c_{2}$ and $f(x)= g^{'}(x)h^{'}(x)$.
So, $g^{'}(x),\ h^{'} \in Z[x]$ and $deg\ g_{'}(x) = deg\ g(x)$ and $deg\ h_{'}(x) = deg\ h(x)$

$$\therefore \text{Reducibility over }Q\text{ impiles reducibility over }Z.$$