*Before the question, I clarify that by "polynomial", I mean any function in the form of
$f(x) = \sum c_ix^i$.
I am wondering whether there exist a polynomial which is not constant (so that it has some x with non-zero factor) but satisfies
$ f(1) = f(2) = 0 $
I intuitively think there is no such polynomial but have no clue how to prove that.
On
José's answer is perfect. But because you asked in the comments, let me perhaps say a bit more.
If a polynomial $p(x)$ has a root $a$, then $x-a$ is a factor of $p(x)$
Proof. We can long divide polynomials. So long divide $p(x)$ by $x-a$, and hence write $p(x)= (x-a)q(x) + r(x)$ for some polynomials $q(x)$ and $r(x)$. Now this equality holds for any value of $x$. But evaluating at $x=a$, we find $p(a)=0= 0+r(x)=r(x)$. Hence, $r(x)=0$. But then $p(x)= (x-a)q(x)$, i.e. $x-a$ is a factor of $p(x)$.
So if you want a polynomial with roots $x=1,2$, then $x-1$ and $x-2$ will both be factors of the polynomial. So your polynomial must be of the form $(x-1)(x-2)q(x)$ for some polynomial $q(x)$. To find one of the simplest possible examples, take $q(x)=1$, then $$ (x-1)(x-2)= x^2-3x+2, $$ as in José's beautifully simple answer. Of course, there are infinitely many examples: $$ \begin{split} (x-1)(x-2)x&= x^3-3x^2+2x \\ (x-1)(x-2)(x-1)&= x^3-4x^2+5x-2 \\ (x-1)(x-2)(x^2+1)&= x^4-3x^3+3x^2-3x+2 \\ \end{split} $$ Generally, if you want a polynomial with (for simplicity, let's say distinct roots) $a_1,a_2,\ldots,a_n$, then $$ p(x)= (x-a_1)(x-a_2)\cdots(x-a_n)q(x), $$ will work, where $q(x)$ is any polynomial. Note that $a_1,\ldots,a_n$ may not be all the roots because, depending on your choice of $q(x)$, $q(x)$ (and hence $p(x)$) may have roots. As a final note, if you wanted these to be all the roots, e.g. in the example above, you had better choose $q(x)$ to be a polynomial with no real roots, as in my last example above.
What about $f(x)=(x-1)(x-2)=x^2-3x+2$?