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Let $p$ be a polynomial of degree $n$. Prove that it has at most n zeros.

Use induction and mean value theorem.

I don't understand how to do the induction. I used $n=0$ for the base case which is obvious and then I assumed that for degree $n$ that the polynomial had at most n roots. Then I was trying to prove that for $n+1$ it had at most $n+1$ roots.

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  • $\begingroup$ I used n=0 for the base case which is obvious and then I assumed that for degree n that the polynomial had at most n roots. Then I was trying to prove that for n+1 it had at most n+1 roots. im in general just lost. $\endgroup$ – Johan Apr 30 '15 at 17:10
  • $\begingroup$ Where did you get stuck? Making that assumption is the easy part $\endgroup$ – jameselmore Apr 30 '15 at 17:12
  • $\begingroup$ I got stuck on how to use mvt to show this. $\endgroup$ – Johan Apr 30 '15 at 17:13
  • $\begingroup$ But you can't assume that a polynomial degree n+1 has n+1 roots and I'm pretty sure that's where you use rolles theorem but I don't understand how it works out. $\endgroup$ – Johan Apr 30 '15 at 17:15
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Hint: (I'll let you formalize it)
Let $g(x) \in P_{n+1}$ be a polynomial from the set of all polynomials of degree $n+1$ ($P_{n+1}$).
Assume that $g$ has $k > n+1$ distinct roots.

Clearly, $g'(x)\in P_n$, and since $g$ has $k$ distinct roots, there are $(k-1)$ points where $g'(x) = 0$. However, $k - 1 > (n+1) - 1 > n$, and we come to a conclusion the $g'(x)$ has $k-1 > n$ roots...

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  • $\begingroup$ Just to point it out: the mean value theorem is used to prove that $g'$ has $k-1$ distinct roots. $\endgroup$ – A.P. Apr 30 '15 at 18:03

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