# Content of Rational Polynomial is Multiplicative/Proof 1

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## Theorem

Let $h \in \Q \sqbrk X$ be a polynomial with rational coefficients.

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

Then for any polynomials $f, g \in \Q \sqbrk X$ with rational coefficients:

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

## Proof

Let $f^* = \dfrac 1 {\cont f} f$, $g^* = \dfrac 1 {\cont g} g$

By Content of Scalar Multiple:

- $\cont {f^*} = \cont {g^*} = 1$

That is, $f^*$ and $g^*$ are primitive.

By Gauss's Lemma on Primitive Rational Polynomials, it follows that $f^* g^*$ is primitive.

Then:

\(\ds \cont {f g}\) | \(=\) | \(\ds \cont f \cont g \cont {f^* g^*}\) | Content of Scalar Multiple | |||||||||||

\(\ds \) | \(=\) | \(\ds \cont f \cont g\) | because $\cont {f^* g^*} = 1$ |

$\blacksquare$