Analysis involves a polynomial with mult degree

Q: Let $\displaystyle p: \Re \mapsto \Re $be a polynomial with degree no greater than 5.

Suppose that at some point $\displaystyle x_{0} \epsilon \Re$ $\displaystyle p(x_{0}) = p\prime (x_{0}) = . . . = p^{(5)}(x_{0}) = 0.$

Prove that $\displaystyle p(x) = 0 $ $\displaystyle \forall x \epsilon \Re$

My answer so far:

By a theorem that I learned in class today, I have -

Proof: Since $\displaystyle \Re$ is an open interval, and $\displaystyle \exists x_{0} \epsilon\Re$ with $\displaystyle p^{(n)}(x_{0})=0$, therefore $\displaystyle \exists z$ between $\displaystyle x_{0}$ and $\displaystyle x$ such that $\displaystyle p(x) = \frac{p^{(5)}(z)(x-x_{0})}{5!}$

Now, I don't know how to prove $\displaystyle p^{(5)}(z)(x-x_{0})$ equals to zero.

Am I on the right path?

Thank you.

KK