proof of factor theorem using division

Lemma (cf. factor theorem).

Let $R$ be a commutative ring with identity and let $p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in }R\mathit{}\mathrm{[}x\mathrm{]}$ be a polynomial with coefficients in $R$. The element $a\mathrm{\in }R$ is a root of $p\mathit{}\mathrm{(}x\mathrm{)}$ if and only if $\mathrm{(}x\mathrm{-}a\mathrm{)}$ divides $p\mathit{}\mathrm{(}x\mathrm{)}$.

Proof.

Let $p(x)$ be a polynomial in $R[x]$ and let $a$ be an element of $R$.

1. First we assume that $(x-a)$ divides $p(x)$. Therefore, there is a polynomial $q(x)\in R[x]$ such that $p(x)=(x-a)\cdot q(x)$. Hence, $p(a)=(a-a)\cdot q(a)=0$ and $a$ is a root of $p(x)$.
   

2. Assume that $a$ is a root of $p(x)$, i.e. $p(a)=0$. Since $x-a$ is a monic polynomial, we can perform the polynomial long division (http://planetmath.org/LongDivision) of $p(x)$ by $(x-a)$. Thus, there exist polynomials $q(x)$ and $r(x)$ such that:

   $$p(x)=(x-a)\cdot q(x)+r(x)$$

   and the degree of $r(x)$ is less than the degree of $x-a$ (so $r(x)$ is just a constant). Moreover, $0=p(a)=0+r(a)=r(a)=r(x)$. Therefore $p(x)=(x-a)\cdot q(x)$ and $(x-a)$ divides $p(x)$.

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