1. ## Really Easy Question about Zeroes of a Polynomial

Let $\displaystyle \mathbb{F}$ be a field, let $\displaystyle f \in \mathbb{F}[x]$ be a polynomial, and let $\displaystyle \alpha \in \mathbb{F}$.

Then I understand that if $\displaystyle (x-\alpha) | f$, then $\displaystyle f(\alpha) = 0$, this is obvious.

But it doesn't seem obvious to me that if $\displaystyle f(\alpha)=0$, then $\displaystyle (x-\alpha)|f$. How can I show this? Is there some result about factorisation that implies this?

Any help would be appreciated.

2. Since

$\displaystyle \mathbb{F}[x]$

is an integral domain we can use the division algorithm.

Now lets divide

$\displaystyle f(x) \text{ by } (x-\alpha)$

This gives

$\displaystyle f(x)=p(x)(x-\alpha)+r(x)$

but since

$\displaystyle (x - \alpha)$

Is degree 1 r(x) must be degree zero or a constant this gives

$\displaystyle f(x)=p(x)(x-\alpha)+R$

Now just evaluate at alpha

$\displaystyle f(\alpha)=p(x)(\alpha-\alpha)+R=R$

So

$\displaystyle (x - \alpha)|f(x) \iff f(\alpha)=0$

3. for any polynomial we can write:

f(x) - f(a) = (x - a)q(x) (see the previous post).

if f(x) = (x - a)k(x), we get f(a) = (x - a)(k(x) - q(x)).

the LHS is a constant polynomial, and the RHS is a polynomial of degree ≥ 1, which is a contradiction....unless k(x) = q(x),

in which case we have f(a) = 0.