What if a polynomial is identically zero? From Barbeau's Polynomials:

  
*
  
*(a) Is it possible to find a polynomial, apart from the constant $0$ itself, which is identically equal to $0$ (i.e. a polynomial
  $P(t)$ with  some nonzero coefficient such that $P(c)=0$ for each
  number $c$)?
  

And then I thought about 2 hypothesis:


*

*P(c-c)

*I thought about a polynomial such as $ax^2+bx+c=0$, then I could make a polynomial with $a=1$, $b=-x$, $c=0$ which would render $x^2+(-x)x=0$. I tested it on Mathematica with values from $-10$ to $10$ and it gave me $0$ for all these values.


When I went for the answer, I've found:
               
When I went to the answer, I couldn't understand it, can you help me? I'm trying to know what's he doing in this answer, I guess it's a way to prove it, but It's still intractible to me. You can explain me or recommend me some thing for reading. I'll be happy if you also tell me something about my hypothesis. Thanks.
 A: I guess there is no way to forbid people to present contorted arguments to prove a simple result, but they do the reader a disservice by masking the essential point. Here the essential point is that we want a polynomial that vanishes in more points than its degree, and this can only be achieved by the zero polynomial. (In abstract algebra one should specify: polynomial in one variable over a commutative domain, but that is the case here, so you can forget this.) This is (more or less) the proof referred to at the end of your citation, with reference to the Factor Theorem: if $P$ vanishes in distinct points $a_1,a_2,\ldots,a_n$ then $P$ is divisible by the product $(x-a_1)(x-a_2)\cdots(x-a_n)$ (using an induction argument: $P$ is divisible by $x-a_n$ and $P/(x-a_n)$ still vanishes in $a_1,a_2,\ldots a_{n-1}$), so either $P=0$ or $P$ has degree at least $n$. Add to that the fact that there are more numbers one can take for the $a_i$ (namely infinitely many) than the degree of any fixed  polynomial, and you are done.
This argument also immediately shows that one should not expect this result if one only has finitely many numbers to take for the $a_i$, as happens when working over a finite field, as the product taken over all possible values for $a_i$ gives a counterexample. For instance $x(x-1)=x^2-x$ is a nonzero polynomial over $\mathbf Z/2\mathbf Z$ that vanishes "everywhere", i.e., on $\{0,1\}$. Those who have a copy of Lang's Linear Algebra at hand should check how he swindles to establish the result in arbitrary fields, on which he even bases his definition of polynomials; a fine example of a textbook blunder*.
So we are left with the Factor Theorem. But in spite of the capitals, it is no big deal; the text in the question in fact already uses it for $a=0$. But the general case is similar: if $a$ is a value at which a polynomial $P$ in $x$ vanishes, write the $P$ as a polynomial in $y=x-a$ by using $x=y+a$ and expanding; now since $P$ vanishes at $y=0$ the result has a zero constant term, so it is divisible by $y$ QED.

*Upon rereading that book, I think I found that the explanation lies in the fact that in it Lang defines the term field to mean sub-field of $\Bbb C$, for which limited case the result is true. I still think that, even when writing for a specific audience, one should not so constrain established mathematical terms (the same goes for defining "polynomials" as polynomial functions), any in any case avoid talking about "arbitrary fields" if one does.
A: I don't know if this will answer all your questions, but what the author is doing is to assume there is a non-zero polynomial which evaluates to zero at all points, then reach a contradiction. He does this by analyzing the hypothetical polynomial in a number of different ways:

Assume the polynomial $p(x)$ evaluates to $0$ at all points, but is not equal to zero. Then it has some definite, positive degree $n$. Since $p(0) = 0$, we must have $p(x) = x\cdot q_0(x)$ for some non-zero polynomial $q_0$.
Since $p(1) = 0$, we must have $q_0(1) = 0$, so $q_0(x) = (x-1)q_1(x)$. Therefore we must have $p(x) = x(x-1)q_1(x)$ for some non-zero polynomial $q_1$.
Now he continues this until he reaches $p(x) = x(x-1)(x-2)(x-3)\cdots (x-n + 1)(x-n)q_n(x)$ for some polynomial $q_n$. But the left side has degree $n$, while the right side has degree at least $(n+1)$, which is a contradiction.

The linear equation he mentions goes like this:
Assume $p(x)$ has positive degree $n$. I'm going to illustrate with $n = 2$, but it's the same for all positive integers. We can then write $p(x) = ax^2 + bx + c$. To find $a, b$ and $c$, we make use of the following:
$$
0 = p(0) = a\cdot 0^2 + b\cdot 0 + c = c \quad\Longrightarrow \quad c=0
$$
as well as
$$
0 = p(1) = a\cdot 1^2 + b\cdot 1 + c = a + b\quad\Longrightarrow \quad a = -b
$$
and
$$
0 = p(2) = a\cdot 2^2 + b\cdot 2 + c = 4a + 2b \quad\Longrightarrow \quad 2a = -b
$$
to which the only solutions are $a = b = c = 0$, and $p$ is the zero polynomial.

The last one he mentions, Taylor's theorem, depends on the fact that many functions are completely determined by its value and all its derivatives at a single point. Polynomials are among these functions. Since $p$ always evaluates to zero, all its derivatives do too, and the only Taylor series which satisfies this is the Taylor series for the zero function.

Edit: Somehow I completely misread the first paragraph. What he does here is a variation on the first explanation I gave, only he proves that all the coefficients has to be zero the following way:
Assume $p(x) = a_nx^n + \cdots + a_1x + a_0$. Since $p(0) = 0$, we know that $a_0 = 0$. Now examine the function $a_nx^{n-1} + \ldots + a_2x + a_1 = \frac{p(x)}{x}$, and suppose $a_1 \neq 0$. Then the inequalities listed in the text arrives at a contradiction, so $a_1 = 0$. Now proceed similarily with $\frac{p(x)}{x^2} = a_nx^{n-2} + \ldots + a_3x + a_2$, and $a_2$ has to be zero. And so on.                 
