# Content of Scalar Multiple

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## Theorem

Let $f$ be a polynomial with rational coefficients.

Let $\cont f$ denote the content of $f$.

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

Then:

- $\cont {q f} = q \cont f$

## Proof

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

Let $\Z \sqbrk X$ denote the ring of polynomials over $\Z$.

Let $n \in \Z$ such that $n f \in \Z \sqbrk X$.

Then we have:

- $b n \paren {q f} = a n f \in \Z \sqbrk X$

By the definition of content, and using that $a \in \Z$:

- $\cont {b n q f} = a \cont {n f}$

By definition of content:

- $\cont {q f} = \dfrac 1 {b n} \cont {b n q f}$

Combining the above with the definition of $\cont f$:

- $\cont {q f} = \dfrac a b \dfrac 1 n \cont {n f} = q \cont f$

$\blacksquare$