# 2020-03-07

## Theorem 2.21
> If $f(x)\in\mathbb{F}[x]$ and $c\in\mathbb{F}$, then $(x-c)$ is a factor of $f(x)$ if and only if $c$ is a root of $f(x)$.

### $\Rightarrow$
Assume $(c-x)$ is a factor of $f(x)$ then $f(x)=(x-c)q(x)$ then $f(c)=(c-c)q(c)=0$ thus c is a root.
### $\Leftarrow$
Assume $c$ is a root of $f(x)$, then apply division algorithm to find
$$f(x)=(x-c)q(x)+r(x)$$
for some $q,r\in\mathbb{F}[X]$ , with $\deg(r)<\deg(x-c)$, so $r\in\mathbb{F}$, then consider $f(c)$
\begin{align}
f(c)=&(c-c)q(c)+r&=0\\
&0+r&=0
\end{align}
Thus $r=0$, and so $f(x)=(x-c)q(x)$ and we conclude that $(x-c)$ is a factor of $f(x)$.