<s>[INST] not understanding identity theorem for polynomials 
Suppose $f(x)$ is a polynomial such that $f(x) = a_nx^n +
 a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots +a_1 x + a_0$ and there are
at least $n+1$ different values of $x$ for which $f(x) =0$.
(a) If $a_n\neq 0$, then what does the Fundamental Theorem of Algebra
tell us about the number of different values of $x$ for which $f(x) =
 0$?
(b) Why must we have $f(x) = 0$ for all values of $x$?

It is with the answer to question (b) that I have the most difficulty. The answer to (a) and (b) is:

If any of the $a_i$ are nonzero, then $\deg f \le n$. Therefore, by
the Fundamental Theorem of Algebra, if any of the $a_i$ are nonzero,
then $f(x)$ cannot have more than $n$ roots. However, we are told that
there are at least $n+1$ values of $x$ for which $f(x) = 0$. So, it is
impossible for any of the $a_i$ to be nonzero (since then the
Fundamental Theorem of Algebra would force $f(x)$ to have no more than
$n$ roots). Therefore, all the $a_i$ must be 0, which means $f(x) = 0$
for all values of $x$. □

I'm having difficulty understanding what is meant with any of the $a_i$ are nonzero. What does this mean? That none of them are 0?
As far as I'm understanding, $f(x) = 0$, like, literally.
Then, there is the following, related bit:

Suppose that $f(x)$ is a quadratic polynomial such that $f(2) = 4$,
$f(3) = 9$, and $f(4) = 16$. Prove that $f(x)=x^2$.
By inspection, we note that $f(x) = x^2$ satisfies $f(2) = 4$, $f(3) =
 9$, and $f(4) = 16$. However, maybe there are other quadratics $f(x)$
such that $f(2) = 4$, $f(3) = 9$, and $f(4) = 16$. How can we tell if
$f(x)= x^2$ is the only possible one?
We let $g(x) = f(x) - x^2$ and focus on $g(x)$, because we know that
2, 3, and 4 are all roots of $g(x)$. Therefore, the Factor Theorem
tells us that $x-2$, $x-3$, and $x-4$ are all factors of $g(x)$. So,
we have $$ g(x) = (x - 2)(x - 3)(x - 4)q(x), $$for some polynomial
$q(x)$. This tells us that either the degree of $g(x)$ is at least 3,
or $q(x)=g(x)= 0$ for all values of $x$. However, because $g(x) = f(x)
 -x^2$ and $f(x)$ is a quadratic, we know that the degree of $g(x)$ cannot be larger than 2. Therefore, we must have $q(x) = g(x)= 0$ for
all $x$. Finally, since $f(x) = g(x) + x^2$, we know that $f(x) = x^2$
for all $x$.

But why is $g(x) = f(x) - x^2$? I mean, I know why, but why not any other polynomial? In other words, I'm not convinced; I'm still thinking that there might be some polynomial out there which satisfies the conditions.
For example:

$f(1) = 1$, $f(2) = 8$, $f(3) = 27$, and $f(4) = 64$.
This looks like it might be $x^3$, but actually, it is $f(x) = x^3 + (x - 1)(x - 2)(x - 3)(x - 4) \left( 16,\!000,\!000x^{427} 473.15x^{101} - \frac{\pi}{\sqrt{2}} x^{23} + x^5 + 99 \right)$

I could have gone and done:
$g(x) = f(x) - x^3$ with $f(x) = x^3$. Then, $g(x) = (x - 1)(x - 2)(x - 3)(x - 4)q(x)$. Since the degree of $g(x)$ cannot be larger than 3, we must have that $q(x) = g(x) = 0$. Finally, we have $f(x) = g(x) + x^3$, and we know that $g(x) = 0$, and so $f(x) = x^3$.
But apparently it doesn't proof anything because there's a polynomial out there that produces the same result! So what did I proof?
 [/INST] A nonzero polynomial of degree $d$ has at most $d$ roots. And the expression
$$\sum_{k=0}^n a_kx^k$$
is a polynomial of degree at most $n$. Hence if it has $n+1$ roots, it must be identically $0$.

Then if there are two polynomials of degree at most $n$, let $p(x)$ and $q(x)$, achieving the same value at $n+1$ points, by the above theorem, the polynomial $p(x)-q(x)$ has $n+1$ roots and so it must be identically $0$. In other words, $p(x)=q(x)$.
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