Little Bézout Theorem

Theorem
Let $P_n \left({x}\right)$ be a polynomial of degree $n$ in $x$.

Let $a$ be a constant.

Then the remainder of $P_n \left({x}\right)$ when divided by $x-a$ is equal to $P_n \left({a}\right)$.

Proof
By the process of Polynomial Long Division, we can express $P_n \left({x}\right)$ as:
 * $(1) \qquad P_n \left({x}\right) = \left({x-a}\right) Q_{n-1} \left({x}\right) + R$

where:


 * $Q_{n-1} \left({x}\right)$ is a polynomial in $x$ of degree $n-1$;


 * $R$ is a polynomial in $x$ of degree no greater than $0$; that is, a constant.

It follows that, by setting $x = a$ in $(1)$, we get $P_n \left({a}\right) = R$.

Hence the result.