Zero Polynomials: Help Me Get out of a Circular Argument The following exercise appears at the beginning of Apostol's Calculus (Vol. 1), before any calculus has been developed, so it's not looking for a solution involving derivatives or Taylor's theorem:

Let $f(x)$ be a polynomial of degree $n$. If $f(x) = 0$ for $n+1$ distinct real values of $x$, then every
  coefficient $c_k$ is zero and $f(x)=0$ for all real $x$.

I can easily prove the second part - that is, $f(x)=0$ everywhere - by applying the factor theorem $n$ times, but this doesn't say anything about the coefficients. So I tried using an inductive argument, but I couldn't make it work, for two reasons:


*

*If $f(x)$ has degree $n+1$ and is zero everywhere, then $f(x) = xg(x)$ where $g(x)$ has degree $n$. We can say that $g(x)$ is zero for all $x\neq 0$, but not for $x=0$. Therefore we can't apply the induction step.

*Even if we got past problem 1., all we'd have shown is that $f(x)$ is equal to a polynomial of degree $n+1$ with all-zero coefficients. The argument is still not complete, because we don't know that every polynomial has a unique representation. The only way I know how to prove this uses the fact that if a polynomial is zero everywhere, it has all-zero coefficients - which is what we're trying to prove in the first place.
I found a proof that an everywhere-zero polynomial has all-zero coefficients, but this can't possibly be what Apostol is asking for from a beginning calculus student. Is there another, simpler proof?
 A: The key thing it seems you're missing is that the factor theorem is a statement about formal polynomials, not just about the values of polynomial functions.  That is, if $f(x)$ is a polynomial and $f(a)=0$, then there exists a polynomial $g(x)$ such that $$f(x)=(x-a)g(x)$$ and this equation means the two sides are equal as formal expressions, not just that they are equal for all values of $x$.  By "equal as formal expressions" I mean that if $f(x)$ is the polynomial $\sum_{i=0}^n b_i x^i$ and $g(x)$ is the polynomial $\sum_{i=0}^{n-1} c_ix^i$, then $b_i=c_{i-1}-ac_i$ for $i=1,\dots,n$ and $b_0=-ac_0$.  That is, the coefficients of $f(x)$ are what you get by formally multiplying out $(x-a)g(x)$ and combining terms with the same power of $x$.  Indeed, if you perform polynomial long division to find the coefficients of $g(x)$ (which is how you prove the factor theorem), you are exactly solving for numbers $c_i$ which make the equations $b_i=c_{i-1}-ac_i$ true.
Thus when you iterate this and get that $$f(x)=(x-a_1)\cdots(x-a_n)h(x)$$ where $h(x)$ has degree $0$, this is also an equality of formal polynomials: the coefficients of $f(x)$ are what you get by expanding out the product on the right-hand side.  Since $h(a_{n+1})=0$, this means $h(x)$ is the zero polynomial: all its coefficients are $0$.  This means that when you expand out the right-hand side, all the coefficients are $0$, so all the coefficients of $f(x)$ are $0$.
Alternatively (but essentially equivalently), you can prove by induction on degree that any polynomial with infinitely many zeroes must be the zero polynomial.  Given such a polynomial $f(x)$, let $a$ be one of the zeroes, and write $f(x)=(x-a)g(x)$ (an equation of formal polynomials!).  Since $g(x)$ has lower degree but also has infinitely many roots, all its coefficients must be zero by induction on degree, and so all the coefficients of $f(x)$ are zero as well.

To remove any doubt, here's the complete argument, without any mention of the word "polynomial" to avoid any confusion.
Theorem: Let $(b_0,b_1,\dots,b_n)$ be a finite sequence of real numbers ($n\geq 0$).  Suppose there exist $n+1$ distinct numbers $x\in\mathbb{R}$ such that $\sum_{i=0}^n b_ix^i=0$.  Then $b_i=0$ for all $i$.
Proof: We use induction on $n$.  The case $n=0$ is trivial, since $\sum_{i=0}^nb_ix^i$ is just $b_0$ for all $x$, and so if there exists a value of $x$ such that this is $0$, $b_0=0$.
Now suppose the result is true for $n-1$, and let $(b_0,\dots,b_n)$ be such that $\sum_{i=0}^nb_ix^i=0$ for $x=a_0,\dots,a_n$, where the $a_i$ are distinct.  Define a sequence $(c_0,\dots,c_{n-1})$ by descending recursion as follows.  Start with $c_{n-1}=b_n$. Given $c_i$, define $c_{i-1}=b_i+a_0c_i$.  Let $r=b_0+a_0c_0$. [If this seems mysterious, all we are doing here is long division of $\sum_{i=0}^n b_ix^i$ by $x-a_0$: the $c_i$ are the coefficients of the quotient and $r$ is the remainder.]
Now observe that for any $x\in\mathbb{R}$, $$\sum_{i=0}^nb_ix^i=(x-a_0)\sum_{i=0}^{n-1}c_ix^i+r.$$  Indeed, if you expand out the right-hand side using the definitions of $c_i$ and $r$, you get that everything cancels except terms $b_ix^i$ for each $i=0,\dots,n$.  Now if we set $x=a_0$, the left-hand side is $0$, and the right-hand side is $r$, so $r=0$.  Moreover, if we set $x=a_i$ for $i>0$, the left-hand side is still $0$, and the only way the right-hand side can be $0$ is if $\sum_{i=0}^{n-1}c_ix^i=0$.
Thus the sequence $(c_0,\dots,c_{n-1})$ has the property that $\sum_{i=0}^{n-1}c_ix^i=0$ for $x=a_1,\dots,a_n$. By the induction hypothesis, this means $c_i=0$ for all $i$.  
Since $b_n=c_{n-1}$, this means $b_n=0$.  For $0< i<n$, we have $b_i=c_{i-1}-a_0c_i$, so $b_i=0$.  Finally, we have $b_0=r-a_0c_0=0$ since $r=0$ and $c_0=0$.  Thus $b_i=0$ for all $i$, as desired.
(As you alluded to, there are also many analytic ways to prove this result.  But the proof above is purely algebraic, and in fact works with any integral domain in place of $\mathbb{R}$.)
A: If you know linear algebra, you can use Vandermonde determinant.
Let $f(x)=a_0+a_1x+\cdots+a_nx^n$, you want to solve for $a_0, a_1, \cdots, a_n$, knowing that $f$ has $n+1$ roots: $$f(x_0)=0, f(x_1)=0, \cdots, f(x_n)=0$$
Then we get $$
\begin{bmatrix}
1&x_0&\cdots&x_0^n\\
1&x_1&\cdots&x_1^n\\
\cdots &\cdots &\cdots &\cdots \\
1&x_n&\cdots&x_n^n
\end{bmatrix}
\begin{bmatrix}
a_0\\
a_1\\
\cdots\\
a_n\end{bmatrix} = 
\begin{bmatrix}0\\0\\\cdots\\0\end{bmatrix}
$$
This has nontrivial solution iff the determinant of the matrix on the left has determinant $0$. But that's the Vandermonde determinant! So this has nontrivial solution iff
$$\prod_{0\leq i<j\leq n}(x_j-x_i)=0$$
iff $x_i=x_j$ for some $0\leq i<j\leq n$, that is, two of the $n+1$ roots are equal. Since the $n+1$ roots are distinct, we have $a_0, \cdots, a_n = 0$, so $f(x)=0$
A: A polynomial function $p(x)=\sum_{j=0}^nA_jx^j$ with deg $(p)=n> 0$ must have $A_n\ne 0$. And then $p(x)$  cannot be $0$ for all $x\in \mathbb R.$ 
For $x\ne 0$ we have $$p(x)=A_nx^n\left(1+\sum_{j=0}^{n-1}T_j(x)\right)$$ where $T_j(x)=A_n^{-1}A_jx^{-(n+1-j)}.$
Now each $|T_j(x)|,$ (for $j=0,...,n-1$), can be as small as desired,  by taking $|x|$ large enough.  So if $|x|\ne 0$ and  $ |x|$ is large enough that each $|T_j(x)|<1/2n$ then $$|p(x)|\geq |A_n x^n|\cdot (1-\sum_{j=0}^{n-1} |T_j(x)|)\geq \frac {1}{2}|A_nx^n|>0.$$
Remarks. (1). I used the notation $T_j(x)$ to save  on typing.... (2). We must use some property of $\mathbb R$ other than just being a field. For example, in the finite field $\{0,1\}$ the function $p(x)=x^2+x$ is always $0$ but the co-efficients are not all $0.$  
A: The question neglects to mention the previous part of the exercise: 

[The exercise begins by saying let $f$ be a polynomial of degree $n$.]
c) 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 $n-1$.

This makes it clear that the root-factor technique is intended to be used to answer your question.
As someone has already pointed out, the statement of this part of the exercise is not the best: it asks to show that a polynomial of degree $n$ with $n+1$ roots must be the zero polynomial.  But the zero polynomial doesn't have degree $n$ for any natural number $n$ (if you want to give it a degree, the best choice is $-\infty$), so a less convoluted statement would be:

A degree $n$ polynomial has at most $n$ real roots.

Indeed this follows easily from the previous part: suppose $f$ has roots at the distinct numbers $a_1,\ldots,a_n$.  If $f$ has a root at $a_1$, 
then 
$f(x) = (x-a_1)h_{n-1}(x)$,
where $h_{n-1}(x)$ has degree $n-1$.  Now
$0 = f(a_2) = (a_2-a_1)h_{n-1}(a_2)$.
Since $a_2 \neq a_1$, we have $a_2 - a_1 \neq 0$ and thus $h_{n-1}(a_2) = 0$.  Applying the same result to $h_{n-1}$ in place of $f$, we find that there is a polynomial $h_{n-2}(x)$ of degree $n-2$ such that 
$h_{n-1}(x) = (x-a_2)h_{n-2}(x)$
and thus 
$f(x) = (x-a_1)(x-a_2)h_{n-2}(x)$.
Continuing in this way we arrive at the factorization
$f(x) = (x-a_1)(x-a_2) \cdots (x-a_n) h_0(x)$,
where $h_0$ has degree $0$ -- meaning it is a nonzero constant, say $c$.  Thus
$f(x) = c(x-a_1)(x-a_2) \cdots (x-a_n)$.
And now indeed it is clear that $f$ doesn't have any roots other than $a_1,\ldots,a_n$ -- if $b$ is not equal to $a_i$ for any $1 \leq i \leq n$ 
then $c,b-a_1,\ldots,b-a_n$ are all nonzero, so their product is nonzero.
Two final comments:
1) We could shorten the proof and make it more formally correct by applying induction on $n$: if our induction hypothesis is that a degree $n$ polynomial can have at most $n$ distinct roots, then we are already done after writing 
$f(x) = (x-a_1)h_{n-1}(x)$: by induction, $h_{n-1}$ has at most $n-1$ roots, and $f$ has at most those roots together with $a_1$ so at most $n$ roots in all.  However I think it is more natural to actually carry out the procedure and see that it leads to a "complete factorization" for a degree $n$ polynomial with $n$ roots.  
2) I think you were thrown off by the somewhat weird exposition in Barbeau's book (that you linked to).  On the one hand, he seems to want to make use of properties of real numbers and estimates.  That's cute, but as an algebraist I feel that it is somehow missing the point: this result holds for polynomials over an arbitrary integral domain (commutative ring without zero divisors) with essentially the same proof (including the observation that over an arbitrary commutative ring, long division of an arbitrary polynomial by a monic polynomial works as usual).  Moreover the proof shows you why you have to be in  a domain and gives you a clue that working over a finite domain $R$ there may be nonzero polynomials that vanish at every point of $R$.  (There are, and this is important in algebra and number theory.)  On the other hand, Barbeau's phrasing makes it sound like the algebraic argument is not correct!  Of course it is, and it is much more straightforward for a beginning student who has just learned the root-factor theorem [i.e., part c) of the exercise] than the analytic estimates he makes first (though as an instructor of analysis I agree that eventually a student should be able to make this kind of argument as well).
A: Start with the basic Fundamental Theorem of Algebra.
Because each  non-zero polynomial of degree $n$ has $n$ complex roots, by the Theorem, they can be written as 
$\displaystyle k\prod_{c=1}^n(x-r_c)$ 
If we have $n+1$ distinct zeroes in a polynomial supposedly of degree $n$, then this contradicts the Fundamental Theorem of Algebra, and thus, the function must be the zero function.
A: If you have proved that $f(x) = 0$ for all real $x$, you can prove the first part this way:
Let the polynomial be $f(x) = \sum_{i=0}^n c_i x^i$. First consider the case where $c_n > 0$.
For sufficiently large $x > 0$, we have $c_n x^n > \sum_{i=0}^{n-1} |c_i| x^i \ge \sum_{i=0}^{n-1} c_i x^i$, and therefore $f(x) > 0$ which is a contradiction.
If $c_n < 0$, by considering the polynomial $-f(x)$, the same argument shows there is a value of $x$ such that $f(x) < 0$ which is a contradiction.
Therefore $c_n = 0$. By repeating the argument for $c_{n-1}, \dots, c_0$, all the $c_i$ are zero.
