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.