Identifying multiple root of a polynomial from derived polynomial It is known that if $c$ is a root of the real polynomial $f(x)$ of multiplicity $r$, then $c$ is a root of the the derivative of the polynomial $f'(x)$ of multiplicity $(r-1)$.
Does the converse hold in the following sense: If it is known that $c$ is a root of both $f$ and $f'$ having multiplicity $k$ for $f'$, can we conclude that $c$ is of multiplicity $k+1$ for f? 
Added: Can I do it like this: let $c$ be of multiplicity $s$ in $f$.  Then $c'$ is of multiplicity $s-1$ in $f'$. Clearly $s-1=k$ so $s=k+1$.

So we're assuming $f'(x)=(x-c)^kg(x)$ and $f(x)=(x-c)h(x)$, and trying to decide whether $f(x)=(x-c)^{k+1}p(x)$. 
From $f(x)=(x-c)h(x)$ we get $f'(x)=(x-c)h'(x)+h(x)$, so $(x-c)^k$ divides $(x-c)h'(x)+h(x)$, so $x-c$ divides $h(x)$, so $h(x)=(x-c)h_1(x)$, so $f(x)=(x-c)^2h_1(x)$. Then $f'(x)=(x-c)^2h_1'(x)+2(x-c)h_1(x)$, so $(x-c)^k$ divides $(x-c)^2h_1'(x)+2(x-c)h_1(x)$, so (if $k\ge2$) $x-c$ divides $h_1(x)$, so $f(x)=(x-c)^3h_2(x)$. At some point, we get tired of arguing this way, and construct a proof by induction. 
So, here's what we should have done:
Assume $f'(x)=(x-c)^kg(x)$ for some $k\ge1$ and some polynomial $g(x)$.
Assume $f(x)=(x-c)^rh(x)$ for some $r$ with $1\le r\le k$ and some polynomial $h(x)$.
It suffices to prove $f(x)=(x-c)^{r+1}h^*(x)$ for some polynomial $h^*(x)$.
We differentiate $f(x)=(x-c)^rh(x)$ to get $$f'(x)=(x-c)^rh'(x)+r(x-c)^{r-1}h(x)=(x-c)^{r-1}\bigl((x-c)h'(x)+rh(x)\bigr)$$ From $f'(x)=(x-c)^kg(x)$ we get $(x-c)^r$ divides $f'(x)$, and comparing to the displayed formula we see $x-c$ divides $(x-c)h'(x)+rh(x)$, from which it follows that $x-c$ divides $h(x)$.
Thus, $h(x)=(x-c)h^*(x)$ for some polynomial $h^*(x)$, and then $f(x)=(x-c)^{r+1}h^*(x)$. 
