TITLE: Understanding proof of uniqueness in theorem on polynomial interpolation
QUESTION [6 upvotes]: There is a slight part of the following proof in my textbook which I don't quite get.
THEOREM
If $x_0, x_1,...,x_n$ are distinct real numbers, then for arbitrary values $y_0, y_1,...,y_n$, there is a unique polynomial $p_n$ of degree at most $n$ such that
$$p_n(x_i) = y_i \quad (0 \leq i \leq n)$$
PROOF
Let us prove the uniqueness or unicity first.  Suppose there were two such polynomials, $p_n$ and $q_n$.  Then the polynomial $p_n - q_n$ would have the property $(p_n - q_n)(x_i) = 0$ for $0 \leq i \leq n$.  Since the degree of $p_n - q_n$ can be at most $n$, this polynomial can have at most $n$ zeros if it is not the $0$ polynomial.  Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros; it must therefore be $0$.  Hence, $p_n = q_n$.
OK, so the only thing I don't quite here is the part:
"Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros"
How does it follow that this must have exactly $n+1$ zeros.  I have pondered this for some time, but just can't see why this is obvious.  I'm probably overlooking something simple here, but if anyone can help me out, I would greatly appreciate it!

REPLY [2 votes]: You don't know that it has exactly $n+1$ zeros, just that it has at least that many because you have found them:  $x_0, x_1, \ldots, x_n$.  But an $n^{\text{th}}$ degree polynomial can only have $n$ zeros, unless it is identically zero...

REPLY [2 votes]: The phrase "Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros" is intended to mean that there are at least $n+1$ zeros. It would have been a little better to say "at least," but being absolutely precise can be a bit tedious.