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I can see why this works for a root $p$ with multiplicity $k\geq 1$, when $f(x)=(x-p)^k$.

But, why is that true if $f(x)=(x-x_1)(x-x_2)\cdots(x-x_n)$ has distinct roots $x_1\neq x_2\neq \cdots \neq x_n$?

Is it something to do with Lagrange's Mean Value Theorem?

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I guess that instead of "continuous functions" you mean "polynomials"? And by "non-zero derivations" you mean non-identically-null derivatives. If so, and if already know the factorization theorem, then it's almost trivial: just expand the product. –  leonbloy Feb 13 '12 at 20:10
    
this question derives from analysis,i am not asking why the polynom of order n has n roots,it is clear as explained below,i am asking for a proof $f^(k^)(x)\neq0$ for every $1<=k<=n$ –  user24927 Feb 13 '12 at 20:26
    
i know that the the n+1 derivation of $f(x)=a0+a1x+a2x^2+..+anx^n$ & $an\neq0$ will result in zero value,i am looking for a proof based on lagrange mid-value theorem,is there one? –  user24927 Feb 13 '12 at 20:33

2 Answers 2

The statement is not true over all functions and fields, but is certainly true of polynomials over the complex numbers. This is known as the fundamental theorem of algebra.

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yes,i know it is true,i am looking for proof,the fundamental theorem of algebra is irrelevant for the proof as i already given the form $f(x)=(x-x_1)(x-x_2)..(x-x_n)$ –  user24927 Feb 13 '12 at 20:19

Throughout, suppose we're working over a field of characteristic $0$.

Let $$f(x) = \displaystyle \sum_{k=0}^n a_kx^k$$ be a polynomial of degree $n$, so that $a_n \ne 0$. Then its derivative $$f'(x) = \displaystyle \sum_{k=0}^{n-1} (k+1)a_{k+1}x^{k}$$ has leading coefficient $na_n$, which is nonzero if $n>0$. So if $n>0$ then this has degree $n-1$. Apply some sort of induction argument and you'll see that this implies that a polynomial of degree $n$ (over a field of characteristic $0$) has exactly $n$ nonzero derivatives, and that its $(n+1)^{\text{th}}$ derivative is $0$.

So if you're working over $\mathbb{C}$ then a consequence of the fundamental theorem of algebra is that $$\#\text{nonzero derivatives} = \deg f = \#\text{roots}$$

This is true over any algebraically closed field of characteristic $0$. In fact, it's true for a polynomial of degree $n$ over any algebraically closed field of characteristic $0$ or $p>n$.

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