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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.2012-02-13
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    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$2012-02-13
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    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?2012-02-13

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