PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Viewing Version 6 of 'factor theorem'
[ view 'factor theorem' | back to history ]

Title of object: factor theorem
Canonical Name: FactorTheorem
Type: Theorem

Created on: 2002-02-04 01:43:49
Modified on: 2006-06-15 16:15:58

Creator: drini
Modifier: pahio
Author: pahio
Author: drini
Author: alozano

Classification: msc:12D05, msc:12D10
Synonyms: factor theorem=root theorem

Revision comment (for changes between this and next version):

rerendering

Preamble:

%\usepackage{graphicx}
%\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}}
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
$$p(x)=(x+1)(x-5)(x+7).$$