Can a polynomial of $n$ degree have $n+1$ distinct real roots?

Question

Question : Let $f(x) = \sum^n_{k=0}c_kx^k$ be a polynomial function then prove that if $f(x) = 0$ for $n+1$ distinct real values, then every coefficient $c_k$ in $f(x)$ is $0$ , thus $f(x) = 0$ for all real values of $x$.

What I think : My problem is that I have no other original thought as to how come a polynomial function $f(x)$ of degree $n$ have more than $n$ values that satisfy $f(x) = 0$; if we factorize any $n$ degree polynomial we would get something of the form $(x-a_1)(x-a_2)........(x-a_n) = 0$ so we get here $n$ values again if we see graphically, we find that the $f(x)$ curve would meet the $Y$ axis $n$ times.

So does the question mean that $f(x)$ does not exist when it says that all the coefficients in $f(x)$ would be zero if $n +1$ real distinct values satisfy $f(x) = 0$ or does it mean something else and how do I prove it ?

Answer 1

It does exist, it's just $0$. Your factorization is almost right, but you forgot the leading coefficient: it should be $f(x) = c_n (x - a_1) \ldots (x - a_n)$ (where $a_1, \ldots, a_n$ are the first $n$ of those $n+1$ values). Your job is to show that the polynomial has this form, and that $c_n = 0$.

Answer 2

It is actually possible for a polynomial of degree $n$ to have more than $n$ roots. For example, consider the ring $R=\mathbb{Z}/25\mathbb{Z}$. Consider the polynomial $f(x)=x^2\in R[x]$. Then $f$ has $5$ roots, namely, $\overline{0}, \overline{5}, \overline{10}, \overline{15}, \overline{20}$ in $R$.

A polynomial of degree $n$, has $n$ roots if the underlying ring is a field.