Theorem: If $x_0$ is a root of polynomial $P$ with multiplicity $m$, then it's a root of $P'$ with multiplicity of m-1, clarification needed.

Question

So, the supposed in the theorem is that $x_0$ is a root of polynomial $P$ with multiplicity $m$. Then you can express the polynomial $P$ as $P(x)=(x-x_0)^m\cdot Q(x)$, where $Q(x)$ is some polynomial and is not divisible by $x-x_0$. My question is why $Q(x)$ is not divisible by $x-x_0$?

Answer 1

If it was, then we could have written $P(x) = (x-x_0)^{m+1}R(x)$ where $R(x) = \frac{Q(x)}{x-x_0}$. This would contradict root $x_0$ having multiplicity $m$.

Answer 2

Because the meaning of the expression “$x_0$ is a root of $P(x)$ with multiplicity $m$” is that $(x-x_0)^m$ divides $P(x)$ and that furthermore $x_0$ is not a root of the quotient. So, if $Q(x)=\frac{P(x)}{(x-x_0)^m}$, then $x_0$ is not a root of $Q(x)$.

Answer 3

$x_0$ having multiplicity of exactly $m$ means that $x-x_0$ divides $P(x)$ exactly $m$ times - $Q(x)$ had a factor of $x-x_0$ then the multiplicity would be more than $m$.