What is the role of the derivative in the divisibility of polynomials?

Question

So I learnt at school that a polynomial $g=(x-a)^p$ divides a polynomial $f$ only if $f(a)=0, f^{'}(a)=0, ..., f^{(p-1)}(a)=0$ and $f^{(p)}(a)\neq 0$. Now I understand $x-a$ divides $f$ only if $f(a)=0$, $f(a)$ being the remainder of the divide but can you explain me why we use the derivates to show what I said at the beginning? I want to ask my teacher but it is a lot of time until I have math class and I can not wait to find the answer.

Answer 1

Suppose $f$ is a polynomial. We have $f(a)=0$, which means that $f(x)=(x-a)g(x)$.

Now we have $f'(x)=g(x)+(x-a)g'(x)$ and thus $0=f'(a)=g(a)$, which means that $g(x)=(x-a)h(x)$, so we have $f(x)=(x-a)^2h(x)$.

Now this can be repeated up to the $(p-1)$th derivative, so we have $f(x)=(x-a)^p z(x)$, but $f^{(p)}(a)\neq 0$, which means (first write out $f^{(p)}(x)=\frac{d}{dx}(x-a)^pz(x)$) that $z(a)\neq 0$, so $a$ is not a root of $z(x)$. Which means exactly that $(x-a)^p$ divides $f$.