Corollary 23.3

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Statement

(Factor Theorem) An element $a\in F$ is a zero of $f(x)\in F[x]$ if and only if $x-a$ is a factor of $f(x)$ in $F[x]$.

Proof
Suppose that for $a\in F$ we have $f(a)=0$. By Theorem 23.1, there exist $q(x),r(x)\in F[x]$ such that

(1)
\begin{equation} f(x)=(x-a)q(x)+r(x), \end{equation}

where either $r(x)=0$ or the degree of $r(x)$ is less than 1. Thus we must have $r(x)=c$ for $c\in F$,so

(2)
\begin{equation} f(x)=(x-a)q(x)+c. \end{equation}

Applying our evaluation homomorphism,$\phi_a:F[x]\rightarrow F$ of Theorem 22.4$, we find (3) \begin{equation} 0=f(a)=0q(a)+c, \end{equation} so it must be that$c=0$.Then$f(x)=(x-a)q(x)$, so$x-a$is a factor of$f(x)$. Conversely,if$x-a$is a factor of$f(x)$in$F[x]$,where$a\in F$,then applying our evaluation homomorphism$\phi_a$to$f(x)=(x-a)q(x)$, we have$f(a)=0q(a)=0\$.

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