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 $s=k+1$.
 A: 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 $f(x)=(x-c)^{r+1}h^*(x)$. 
A: The answer is yes.   Consider the case when $c=0$.  If $f'$ has a root at $x=0$ of multiplicity $k$, then $f'(x) = x^k g(x)$ where $g$ is a polynomial with degree $n-k\geq0$, and 
$$
f'(x) = \sum_{i=k}^n \alpha_i x^i
$$
for some sequence of real numbers $\{\alpha_i|i=1,2,\ldots, n\}$ with $\alpha_n\neq 0$ and $\alpha_k\neq0$.
Thus by integration,
$$
f(x) = \beta + \sum_{i=k}^n \frac{\alpha_i}{i+1} x^{i+1}
$$ 
where $\beta\in\mathbb R$.
But $f(c)=f(0)=0$, so $\beta=0$ and hence
$$
f(x) = \sum_{i=k}^n \frac{\alpha_i}{i+1} x^{i+1}
$$ has a root at $x=c=0$ of multiplicity $k+1$. 
If $c$ is non-zero, we can apply the same argument.  $f'(x) = (x-c)^kg(x)$, $f'(x) = \sum_{i=k}^n \alpha_i (x-c)^i$, $f(x) = \beta + \sum_{i=k}^n \frac{\alpha_i}{i+1} (x-c)^{i+1}$, $\beta=0$, and thus $f(x)$ has a root at $x=c$ of multiplicity $k+1$.
----------------- Edit --------------------
Actually, I like Jave's proof better.  Suppose that $f(x)$ has a root of degree $s$ at $x=c$ and $f'(x)$ has a root of multiplicity $c'$ at $x=c$.  By the theorem, $c'=s-1$ so $s=c'+1$.  Nice.
