multiplicity of the root of $g(x)=f(x)/f'(x)$, given a root of $f(x)$ and $f'(x)$. Here is a quote from my textbook:

"If $f(x)$ has a root at $x=\alpha$ with multiplicity $m&gt;1$, then $f'(x)$ has a root at $x=\alpha$ with multiplicity $m-1$. Then, the function
\begin{align*}
g(x)=f(x)/f'(x)
\end{align*}
has a root at $x=\alpha$ with multiplicity 1 (a simple root)."

But how is $\alpha$ a root for $g$, since $g(\alpha)=0/0$? I'm quite confused, any help appreciated.
Write $f(x)=(x-a)^m h(x)$, with $h(a)\ne0$. Then
$$
g(x)
= \frac{f(x)}{f'(x)}
= \frac{(x-a)^m h(x)}{m(x-a)^{m-1}h(x)+(x-a)^m h'(x)}
= \frac{(x-a) h(x)}{mh(x)+(x-a)h'(x)}
= (x-a)j(x)
$$
where
$$
j(x)=\frac{h(x)}{mh(x)+(x-a)h'(x)}
$$
Since $j(a)=\dfrac{h(a)}{m h(a)}=\dfrac{1}{m}\ne0$, we get that $a$ is a simple root of $g$.
Write $f(x)=(x-a)^m h(x)$, with $h(a)\ne0$. Then
$$
g(x)
= \frac{f(x)}{f'(x)}
= \frac{(x-a)^m h(x)}{m(x-a)^{m-1}h(x)+(x-a)^m h'(x)}
= \frac{(x-a) h(x)}{mh(x)+(x-a)h'(x)}
= (x-a)j(x)
$$
where
$$
j(x)=\frac{h(x)}{mh(x)+(x-a)h'(x)}
$$
Since $j(a)=\dfrac{h(a)}{m h(a)}=\dfrac{1}{m}\ne0$, we get that $a$ is a simple root of $g$.