How to establish these two facts about polynomials? Let $f(x) := \sum_{k=0}^n c_k x^k $ be a polynomial of degree $n\geq 0$ with real coefficeints such that $f(x) = 0$ for $n+1$ distinct real values of $x$. Then how to prove that each $c_k = 0$ and that $f(x) = 0$ for all real $x$? 
We don't have to use the methods of calculus and have to use only the most elementary ideas. 
And how to prove the following statement without recourse to the division algorithm? 
If $n \geq 1$ and $f(a) = 0$ for some real $a$, then $f(x) = (x-a) h(x)$, where $h$ is a polynomial of degree $n-1$. 
 A: Here's the most elementary way I could think of. I prove the second statement first - suppose $f(x)$ is a polynomial of degree $n$, and $f(a)=0$ for some real number $a$. Then $g(x)=f(x+a)$ is a polynomial of degree $n$, and $g(0)=0$, therefore $g(x)=xh(x)$ where $h(x)$ is a polynomial of degree $n-1$. Now $f(x)=g(x-a)$ completes this statement.
Now suppose $f(x)$ is a degree $n$ polynomial with $n+1$ real roots. Apply the preceeding statement to find that $f$ must also be an $n+1$ degree polynomial:
$$f(x) = (x-a_1)\cdots(x-a_{n+1})\cdot c$$
where the $a_i$ are the roots. Subtract the left from the right, and we see that the only way this is possible is if $c=0$, hence $f(x)=0$.
A: Hint: Use the Mean Value Theorem repeatedly to find out the following: 
The $1$st derivative has $n$ distinct real roots. 
The $2$nd derivative has $n-1$ distinct real roots. 
...
The $n-1$th derivative has $2$ distinct real roots. 
The $n$th derivative has $1$ real root. 
Of course, $f^{(n)}(x) = n!c_n$. Can you see where to go from here?
A: Consider the special case  $a=0$. If $0$ is a root, then easy to see the polynomial has zero as its constant term, and so $x$ [that is $(x-0)$] is a factor. Now  to reduce to this special case do the shift defining $g(x) = f(x+a)$. Now $f$ having a zero at $a$ is the same as $g$   having a zero at zero. Use binomial theorem you can get what you want without using division algorithm.
