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 ?
Best Answer
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.