Content of Scalar Multiple

Theorem
Let $f$ be a polynomial with rational coefficients.

Let $\operatorname{cont}\left({f}\right)$ be the content of $f$.

Let $q \in \Q$ be a rational number.

Then:
 * $\operatorname{cont}\left({q f}\right) = q \operatorname{cont}\left({f}\right)$

Proof
Let $q = a/b$ with $a, b \in \Z$.

Let $n \in \Z$ such that $n f \in \Z \left[{X}\right]$.

Then we have $b n \left({qf}\right) = a n f \in \Z \left[{X}\right]$.

By the definition of content, and using that $a \in \Z$:
 * $(1): \quad \operatorname{cont} \left({b n q f}\right) = a \operatorname {cont} \left({n f}\right)$

By definition of content:
 * $\operatorname{cont} \left({q f}\right) = \dfrac 1 {bn} \operatorname{cont} \left({b n q f}\right)$

Therefore, using $(1)$ and the definition of $\operatorname{cont} \left({f}\right)$:
 * $\operatorname{cont}\left({qf}\right) = \dfrac a b \dfrac 1 n \operatorname{cont} \left({n f}\right) = q \operatorname{cont} \left({f}\right)$