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proof of factor theorem using division
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(Proof)
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Proof. Let $p(x)$ be a polynomial in $R[x]$ and let $a$ be an element of $R$
- 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 $p(x)$
- 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: $$p(x)=(x-a)\cdot q(x) + r(x)$$ 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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"proof of factor theorem using division" is owned by alozano.
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Cross-references: degree, monic polynomial, divides, root, coefficients, polynomial, identity, commutative ring, factor theorem
There is 1 reference to this entry.
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 4297 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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Pending Errata and Addenda
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