Double Root of Polynomial is Root of Derivative

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

Let $f \in R \sqbrk X$ 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:
 * $\map f X = \paren {X - a}^2 \map g X$

with $\map g X \in R \sqbrk X$.

From Formal Derivative of Polynomials Satisfies Leibniz's Rule:


 * $\map {f'} X = 2 \paren {X - a} \map g X + \paren {X - a}^2 \map {g'} X$

and thus:
 * $\map {f'} a = 0$