Double Root of Polynomial is Root of Derivative

Theorem
Let $R$ be a ring.

Let $R \left[{x}\right]$ be the ring of polynomial forms over $R$.

Let $f \in R \left[{x}\right]$.

Suppose that $a \in R$ is a double root of $f$.

Let $f'$ denote the formal derivative of $f$.

Then $a$ is a root of $f'$.

Proof
Because $a$ is a double root of $f$, we can write:
 * $f \left({x}\right) = \left({x - a}\right)^2 g \left({x}\right)$

with $g \left({x}\right) \in R \left[{x}\right]$.

From Formal Derivative of Polynomials Satisfies Leibniz's Rule:


 * $f' \left({x}\right) = 2 \left({x - a}\right) g \left({x}\right) + \left({x - a}\right)^2 g' \left({x}\right)$

and thus:
 * $f' \left({a}\right) = 0$