# proof of factor theorem using division

###### Lemma (cf. factor theorem).

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

###### Proof.

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

1. 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. 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 $p(x)$.

Title proof of factor theorem using division ProofOfFactorTheoremUsingDivision 2013-03-22 15:08:58 2013-03-22 15:08:58 alozano (2414) alozano (2414) 8 alozano (2414) Proof msc 12D10 msc 12D05