# $(x-a)^k|f'(x)$ implies $(x-a)|f(x)$ when all complex roots of polynomial $f(x)$ are real and where $k \geq 2$

suppose all complex roots of polynomial $f(x)$ are real. then $$(x-a)^k|f'(x)\Rightarrow (x-a)|f(x)$$where $k\geq2$

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We need to assume that $f(x)$ has degree $\ge 1$. For in a trivial sense the polynomial $17$ is a counterexample to the assertion.

Let $f(x)$ have degree $n \ge 1$, and let the distinct real roots of $f(x)$, say in increasing order, be $a_1,a_2,\dots, a_k$, with multiplicities $e_1,e_2, \dots,e_k$ respectively.

If all the roots of $f(x)$ are real, then $\sum_{i=1}^k e_i=n$.

By Rolle's Theorem, there is a root of $f'(x)$ strictly between any two distinct roots of $f(x)$. That accounts for $k-1$ roots of $f'(x)$.

It is a standard result about polynomials that $f'(x)$ has a root of multiplicity at least $e_i-1$ at every $a_i$. That accounts for $\sum_{i=1}^{k} (e_i-1)$ roots of $f'(x)$.

Add up. We have accounted for $(n-k)+(k-1)=n-1$ roots of $f'(x)$. There cannot be any more, since $f'(x)$ has degree $n-1$. (Here we have used the assumption that $f(x)$ is non-constant.) In particular, no root of $f'(x)$ that is between consecutive $a_i$ can have multiplicity greater than $1$.

Thus if $f(a)\ne 0$, we cannot have a root of $f'(x)$ of multiplicity greater than $1$ at $x=a$. This completes the proof of the assertion.

Added: As has been pointed out by alex.jordan in a question/comment, under the conditions of the problem we can conclude that $(x-a)^{k+1}$ divides $f(x)$.

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just a doubt: if $\sum_{i=1}^{k} (e_i)=n,$ is not $\sum_{i=1}^{k} (e_i-1)=\sum_{i=1}^{k} (e_i)-k=n-k?$ and $n-k< n-1$ if $k>1$ –  lab bhattacharjee Nov 17 '12 at 6:45
@labbhattacharjee: Remember the $k-1$ "in between" roots that I mentioned earlier (Rolle's Theorem). I have added a little bit saying $(n-k)+(k-1)=n-1$. –  André Nicolas Nov 17 '12 at 6:49
thanks for the explanation. –  lab bhattacharjee Nov 17 '12 at 7:07
+1 Does this in fact show, under the conditions given, that $(x-a)^{k+1}|f(x)$? –  alex.jordan Nov 17 '12 at 7:28
Yes, it is the same counting argument. Alternately, one can set $f(x)=(x-a)g(x)$, differentiate, conclude that $x-a$ divides $g(x)$, and keep pushing up. –  André Nicolas Nov 17 '12 at 8:04