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[parent] proof of factor theorem using division (Proof)
Lemma 1 (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. 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 of $ p(x)$ by $ (x-a)$. Thus, there exist polynomials $ q(x)$ and $ r(x)$ such that:
    $\displaystyle p(x)=(x-a)\cdot q(x) + r(x)$
    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)$.
$ \qedsymbol$



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Cross-references: degree, monic polynomial, divides, root, coefficients, polynomial, identity, commutative ring, factor theorem
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This is version 5 of proof of factor theorem using division, born on 2005-03-22, modified 2006-06-20.
Object id is 6896, canonical name is ProofOfFactorTheoremUsingDivision.
Accessed 2833 times total.

Classification:
AMS MSC: 12D05 (Field theory and polynomials :: Real and complex fields :: Polynomials: factorization)
 12D10 (Field theory and polynomials :: Real and complex fields :: Polynomials: location of zeros )
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