Double Root of Polynomial is Root of Derivative

Theorem
Let $R$ be a commutative ring with unity.

Let $f \in R \left[{X}\right]$ be a polynomial.

Let $a \in R$ be a root of $f$ with multiplicity at least $2$.

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

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

Proof
Because $a$ has multiplicity at least $2$, 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:
 * $f' \left({a}\right) = 0$