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<record version="7" id="1810">
 <title>factor theorem</title>
 <name>FactorTheorem</name>
 <created>2002-02-04 01:43:49</created>
 <modified>2008-08-28 06:40:27</modified>
 <type>Theorem</type>
 <creator id="3" name="drini"/>
 <author id="2872" name="pahio"/>
 <author id="3" name="drini"/>
 <author id="2414" name="alozano"/>
 <classification>
	<category scheme="msc" code="12D05"/>
	<category scheme="msc" code="12D10"/>
 </classification>
 <synonyms>
	<synonym concept="factor theorem" alias="root theorem"/>
 </synonyms>
 <related>
	<object name="Polynomial"/>
	<object name="RationalRootTheorem"/>
	<object name="Root"/>
	<object name="APolynomialOfDegreeNOverAFieldHasAtMostNRoots"/>
 </related>
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%\usepackage{xypic} 
\usepackage{bbm}
\newcommand{\Z}{\mathbbmss{Z}}
\newcommand{\C}{\mathbbmss{C}}
\newcommand{\R}{\mathbbmss{R}}
\newcommand{\Q}{\mathbbmss{Q}}
\newcommand{\mathbb}[1]{\mathbbmss{#1}}</preamble>
 <content>If $f(x)$ is a polynomial over a ring with identity, then $x-a$ is a factor if and only if $a$ is a root (that is, $f(a)=0$).

This theorem is of great help for finding factorizations of higher degree polynomials. As example, let us think that we need to factor the polynomial $p(x)=x^3+3x^2-33x-35$. With some help of the rational root theorem we can find that $x=-1$ is a root (that is, $p(-1)=0$), so we know $(x+1)$ must be a factor of the polynomial. We can write then
$$p(x)=(x+1)q(x)$$
where the polynomial $q(x)$ can be found using long or synthetic division of $p(x)$ between $x-1$. In our case $q(x)=x^2+2x-35$ which can be easily factored as $(x-5)(x+7)$. We conclude that
$$p(x)=(x+1)(x-5)(x+7).$$</content>
</record>
