Reducibility over $\mathbb Q$ implies reducibility over $\mathbb Z$.

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Let $f(x) \in \mathbb Z[x]$. If $f(x)$ is reducible over $\mathbb Q$, then it is reducible over $\mathbb Z$.

I went through the proof from the book I'm reading, which starts as follows:

We're given $f(x)$ is reducible over $\mathbb Q$ , so we can write $f(x) = g(x)h(x)$ where $h(x),g(x) \in \mathbb Q[x]$.

Now the next statement says - "Clearly, we may assume that $f(x)$ is primitive because we can divide both $f(x)$ and $g(x)$ by the content of $f(x)$."

How is that ? When we write $f(x) = g(x)h(x)$, how can we conclude that content of $f(x)$ divides $g(x)$ ? It divides $f(x)$, that's obvious, but $g(x)$ ? How ?

Let's say the content of $f(x)$ divides $g(x)$, then how can we conclude that $f(x)$ is primitive ? Can anyone explain ?