If something holds for n+1 distinct values then it holds for all values.Proving a property of polynomial Ok I am stuck on proving a property of polynomial.It basically goes like this.
If $$f(x) = \sum_{k=0}^{n} c_kx^k $$ is equal to zero for n+1 distinct real values x,then f(x) is equal to 0 for all real x and $c_k = 0 $.
I have tried making a contradiction as in : There exist n+1 distinct real values of x for which f(x) = 0 and $c_k \neq 0 $ .
From there I tried proving a contradiction but I do not not how to proceed,or which polynomial to consider.If this contradiction is to be proven then it would follow that $c_k \neq 0 $ does not hold(because x values exist by hypothesis),and that would conclude it.
This problem comes from Tom Apostol Calculus vol.1 Chapter 1.5 exercise 9 ,subexercise d .
It would be much appreciated if you would offer your help,and if you would instruct me in how to go about proving things of this form(namely if some number of values is satisfy,then all values satisfy).
Thank you in advance  
 A: You can also appeal to the mean value theorem. If $f$ has at least $n+1$ distinct zeroes, then $f'$ has at least $n$ distinct zeroes, $f''$ has at least $n-1$ distinct zeroes, and so on. Thus $f^{(n)}$ has at least one zero. Since $f^{(n)}$ is constant, you get $f^{(n)} \equiv 0$. Thus $f^{(n-1)}$ is constant, and since it has zeroes $f^{(n-1)} \equiv 0$. Keep going until you get $f \equiv 0$.
A: This can be proved simply by induction and the Factor Theorem (FT).  Suppose a polynomial $\rm\, f(x)\,$ has more roots than its degree. We prove by induction on degree $\rm\,f\,$ that all coefficients of $\rm\,f\:$ are $\,0.\,$ If $\ \deg f = 0\,$ then $\rm\,f\,$ is constant $\rm\:f = c.\,$ Since $\rm\,f\,$ has a root,  $\rm\,c = 0.\:$ So all coefficients of $\rm\,f\,$ are $\,0.\,$ Else $\rm\,f\,$ has degree $\ge 1,\:$ so $\rm\,f\,$ has a root $\rm\,r.\,$ By the FT, $\rm\,\ f = (x\!-\!r) g(x)\, $ for a polynomial $\rm\,g(x).\,$ Necessarily $\rm\,g\,$ too has more roots than its degree, since all roots $\rm\,s \ne r\,$ are roots of $\rm\,g\,$ because $\rm\,\color{#c00}{s-r\ne 0,\,\ f(s) = (s\!-\!r)g(s) = 0\, \,\Rightarrow\,g(s)=0}.\,$  Hence, by induction, all coefficients of $\rm\,g\,$ are $\,0,\,$ therefore $\rm\,f = (x\!-\!r)g\: $ has all coefficients $0.\ \ $ QED
The $\rm\color{#c00}{red}$ property is crucial. Rings satisfying $\,ab=0\,\Rightarrow\,a=0\,$ or $\,b=0,\,$ i.e. are called integral domains, or simply domains. 
The proof fails over non-domains, e.g. $\rm\,x^2\!-\!1 = (x\!-\!1)(x\!+\!1)\,$  has $\,4\,$ roots $\,\pm1,\pm3\,$ over $\, \mathbb Z/8.\,$ Notice how the proof breaks down due to the existence of zero-divisors: here $\,3\,$ is a root since $\,2\cdot4\equiv 0,\,$ but $\,3\,$ is not a root of either $\rm\,x\!-\!1\,$ or $\rm\,x\!+\!1;\,$ i.e. $\rm\,x\!-\!3\,$ divides $\rm\,(x\!-\!1)(x\!+\!1)\,$ but doesn't divide either factor, so it is a non-prime irreducible. This yields the nonunique factorization $\rm\,(x-3)(x+3)\equiv (x-1)(x+1).$
A: Every non-zero polynomial of degree $n$ can be factored completely over the complex numbers, as a product of exactly $n$ linear terms $(x - a_i)$ (where some terms might be duplicated). This implies that every non-zero polynomial of degree $n$ has at most $n$ distinct roots. If you have a polynomial of degree at most $n$ with more roots than $n$, then it must be the zero polynomial.
A: An $n$-th degree polynomial has at most $n$ zeroes. Let $p(x) = a_n x^n + \cdots a_0$ be the polynomial, where $a_n \ne 0$.
First, if $p(\alpha) = 0$, then $x - \alpha$ divides $p(x)$. To prove this, notice that:
$$
x^k - \alpha^k = (x - \alpha) (x^{ḱ - 1} + x^{k - 2} + \cdots + 1)
$$
Then:
$$
p(x) = p(x) - p(\alpha) = a_n (x^n - \alpha^n) + \cdots + a_1 (x - \alpha)
$$
Each term is divisible by $x - \alpha$.
Next, the main clain is proved by induction on the degree of $p$.
Basis is that if the degree is 1, there is exactly one zero.
Suppose then that all polynomials of $m$-th degree have at most $m$ zeroes, and take a polynomial $p$ of degree $m + 1$. If it has no zeroes, we are done. If it has a zero $\alpha$, by the previous result $p(x) = (x - \alpha) q(x)$, where $q$ has degree $m$. By induction, $q$ has at most $m$ zeroes, and $p$ has at most $m + 1$.
Finally, a polynomial $f(x)$ of degree $n$ with $n + 1$ zeroes can't exist.  The only possibility is that all coefficients of $f$ are zero.
