> [!NOTE] **Theorem (Factor Theorem)** > Let $F$ be a [[Field (Algebra)|field]]. Let $f$ be a [[Ring of Polynomial Forms|polynomial]] over $F$ in $x.$ Then $\exists \alpha \in F: f(\alpha) \iff \exists q\in F[x]: f = (x-\alpha)q $ *Proof:* ($\implies$) Suppose there exist $\alpha\in F$ such that $f(\alpha)=0$. By [[Division with remainder for integers|division with remainder theorem]], $\exists q,r \in F[x]$ with $\deg(r) < \deg((x-\alpha))$ such that $f(x)=(x-\alpha)q(x)+r(x)$We must have $r(x)= \beta$ for some $\beta \in F$ since $0=\deg(r) < \deg((x-\alpha))=1.$ But $0=f(\alpha)=(\alpha-\alpha)q(\alpha)+\beta= \beta$($\impliedby$) Conversely, if there exists $q(x)\in F(x)$ such that $f(x)=(x-\alpha)q(x)$ then $f(\alpha)=(\alpha-\alpha)g(\alpha)=0$. Moreover by [[Degree of Product of Polynomials Over Integral Domain]], $\deg(q)=\deg(f)-1.$