Zero of a polynomial and divisbility

Question

I have a polynomial $p(x)$, $deg(p) = n$. I know that $\alpha$ is a zero of $p(x)$. Then $(x-\alpha)|p(x)$. Is it wrong to say that $(x-\alpha)^m|p(x)$, $m \in \mathbb{N}, m>1 $?

Answer 1

Yes it's wrong to say so.

Take for example:

$$p(x) = x^3-1$$ $$=(x-1)(x^2+x+1)$$

Now note that $x=1$ is a root of $p(x)$ and hence $x-1$ divides $p(x)$. But $x^2+x+1$ has no real roots.

So the claim is false in general.

Answer 2

Yes, it's too constrained to be arbitrary. One constraint is, $$m<n$$ Because, if you don't, you have to multiply by something that isn't a polynomial at some point. As Matt Parker shows us, the coeffients of a polynomial, depend on the choices of individual terms of its factorization.