general theorem on roots of a polynomial needed to show it's identically zero.

Question

1. Polynomial degree k, one variable, if it's zero at k+1 values, then it's identically zero. Can someone point me to a proof of this?

2. I know derivatives at points can count as these roots (if k-degree f(x) vanishes at x=a and k derivatives vanish at x=a, it vanishes). So, can someone point me to a more general theorem incorporating the derivatives?

3. What if, for example, there's a k degree polynomial (in one variable) that vanishes at x=a, also at x=b, and k-1 derivatives vanish at x=a. This is a total of k+1 "roots"; does this count? What if some of those x=a vanishings are other x values?

Answer 1

Proposition: Let $f$ be a polynomial, and let $a \in \mathbb{R}, n \in \mathbb{N}$. If for each $0 \leq k \leq n-1$ we have $f^{(k)}(a) = 0$, then $$f(x) = (x-a)^n g(x)$$ for some polynomial $g$.

Obs: $g$ may be identically zero (which implies that $f = 0$ as well). Since we either have $f = 0$ or $\deg(g) = \deg(f) - n$, this result should be enough for most of your needs.

Proof (by induction on $n$): The case $n = 0$ is easy. Suppose the result holds for $n-1$, and suppose $f(a) = f'(a) = \ldots = f^{(n-2)}(a) = f^{(n-1)}(a) = 0$. By induction, we know in particular that $f(x) = (x-a)^{n-1}h(x)$, so that $f'(x) = (x-a)^{n-2}((n-1)h(x) + (x-a)h'(x))$.

On the other hand, the polynomial $f'(x)$ also satisfies the hypothesis for $n-1$, so by induction we have $f'(x) = (x-a)^{n-1}q(x)$ for some polynomial $q$. Hence, $$(x-a)^{n-2}((n-1)h(x) + (x-a)h'(x)) = (x-a)^{n-1}q(x), $$ or $(n-1)h(x) = (x-a)(q(x) - h'(x)),$ and thus $h(x) = (x-a)g(x)$ for $g(x) = \frac{q(x) - h'(x)}{n-1}.$ Therefore,

$$f(x) = (x-a)^{n-1}h(x) = (x-a)^ng(x).\tag*{$\blacksquare$}$$