TITLE: Polynomial $p(x) = 0$ for all $x$ implies coefficients of polynomial are zero
QUESTION [13 upvotes]: I am curious why the following is true. The text I am reading is "An Introduction to Numerical Analysis" by Atkinson, 2nd edition, page 133, line 4.
$p(x)$ is a polynomial of the form: 
$$ p(x) = b_0 + b_1 x + \cdots + b_n x^n$$
If $p(x) = 0$ for all $x$, then $b_i = 0$ for $i=0,1,\ldots,n$. 
Why is this true? For example, for $n=2$, I can first prove $b_0=0$, then set $x=2$ to get a linear system of two equations. Then I can prove $b_1=b_2 = 0$. Similarly, for $n=3$, I first prove $b_0=0$, then I calculate the rank of the resulting linear system of equations. That shows that $b_1=b_2=b_3=0$. But if $n$ is very large, I cannot keep manually solving systems of equations. Is there some other argument to show all the coefficients must be zero when the polynomial is always zero for all $x$?
Thanks.

REPLY [10 votes]: Note that $p(x)=0$ implies derivatives of all orders are also $0$. Let $x=0$ into $p(x) = b_0 + b_1 x + \cdots + b_n x^n$ to obtain $b_0 = 0$. 
Now differentiate both sides: $p'(x) = b_1 + 2b_2 x + 3b_3 x^2 + \cdots + nb_n x^{n-1} $. Let $x=0$ again, and we get that $b_1=0$. 
If we continue to differentiate and substitution $x=0$ we will get that $b_k=0$ for $k=0,1,2,\cdots, n$. This idea can easily be made into a rigorous induction proof. 
A nice corollary of this result is that two polynomials are equal if and only if they have the same degree and coefficients.