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

2.
Assume that $a$ is a root of $p(x)$, i.e. $p(a)=0$. Since $xa$ is a monic polynomial^{}, we can perform the polynomial long division (http://planetmath.org/LongDivision) of $p(x)$ by $(xa)$. Thus, there exist polynomials $q(x)$ and $r(x)$ such that:
$$p(x)=(xa)\cdot q(x)+r(x)$$ and the degree of $r(x)$ is less than the degree of $xa$ (so $r(x)$ is just a constant). Moreover, $0=p(a)=0+r(a)=r(a)=r(x)$. Therefore $p(x)=(xa)\cdot q(x)$ and $(xa)$ divides $p(x)$.
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Title  proof of factor theorem using division 

Canonical name  ProofOfFactorTheoremUsingDivision 
Date of creation  20130322 15:08:58 
Last modified on  20130322 15:08:58 
Owner  alozano (2414) 
Last modified by  alozano (2414) 
Numerical id  8 
Author  alozano (2414) 
Entry type  Proof 
Classification  msc 12D10 
Classification  msc 12D05 