Let $p(x)$ be a complex polynomial. Prove that if $a$ is multiple root of $p$ then $a$ is root of $p'(x)$.

Question

Let $p(x)$ be a complex polynomial. Prove that if $a$ is multiple root of $p$ then $a$ is root of $p'(x)$.

Note: $p'(x)$ is the polynomial derived from $p$.

Let m be the multiplicity of $a$ as the root of $p$. Then, $p = (x-a) ^ m h$, where $h$ is a complex polynomial, and $a$ is not a root of $h$. As $m> 1$ then $p '= m (x-a) ^{m-1} h + (x-a) ^ m h'$, where $p'(a)= m (a-a)^{m-1} h + (a-a)^m h'=0$.

Is my demonstration correct? If correct, can you think of another way to demonstrate this statement?

Answer 1

Your demonstration seems okay to me. I would probably choose to factor

$$\begin{split} p '(x)&= m (x-a) ^ {m-1} h(x) + (x-a) ^ m h'(x)\\ &=(x-a)^{m-1}(mh(x) +(x-a) h'(x)) \end{split} $$ This shows directly that $x-a$ is a factor of $p'(x)$ because $m-1>0$ without using the factor theorem.