# Rolle's theorem for showing that $(x-a)^k$ divides $p(x)$ . . .

Have the following I'm stuck on:

Suppose $p(x)=p_0+p_1 x+p_2 x^2+\cdots+p_n x^n$ is a polynomial of degree $n \geq 1$. Show that if $(x-a)^k$ divides $p(x)$ for some $a\in\mathbb R$ and some integer $k \geq 1$, then $(x-a)^{k-1}$ divides the derivative $p'(x)$. [Any standard results must be clearly stated.] Is the converse of this result true? Justify.

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Does the exercise explicitly state that one must use Rolle's theorem? – Bill Dubuque Apr 27 '12 at 15:26

If $p(x)=(x-a)^k q(x)$ then $p'(x)=k(x-a)^{k-1}q(x) + (x-a)^kq'(x) = (x-a)^{k-1}(k q(x) + (x-a)q'(x))$.
For the converse, write $p(x)=(x-a)^k q(x) + r(x)$, with $r$ of degree less than $k$. Then, by the previous computation, $(x-a)^{k-1}$ divides $p'(x)$ iff $(x-a)^{k-1}$ divides $r'(x)$. Since $r'(x)$ has degree less than $k-1$, this can only happen when $r'=0$. This gives you a counter-example for the converse: $p(x)=(x-a)^k+1$.
You don't actually need Rolle's Theorem for this. Hint: Write $p(x) = (x-a)^k q(x)$, differentiate using the product rule and conclude the statement you wish to prove. About the converse: Think about constant functions.