Home
Math by Grades Pre-K
Kindergarten
Grade 1
Grade 2
Grade 3
Grade 4
Grade 5
Grade 6
Grades 7 and 8
Grades 9 and 10
Grades 11 and 12
Math by Topics Arithmetic
Algebra
Geometry
Math Word Problems
Trigonometry
Statistics
Probability
PreCalculus
Calculus
Set Theory
Matrices
Vectors
Math Worksheets Math Worksheets
Interactive Zone
Math in Video Lessons Basic Algebra
Intermediate Algebra
College Algebra
High School Geometry
College Calculus
Linear Algebra
Engineering Math
Singapore Math
Math for Specific Tests SAT Math
ACT Math
GMAT Math
High School, Regents
California Standards
GCSE Maths
A Level Maths
Math Fun and Games Math Trivia
Math Games
Fun Games
Mousehunt Guide
Exam Preparation SAT Preparation
ACT Preparation
GMAT Preparation
Science Biology
Chemistry
Science Projects
High School Biology
High School Chemistry
High School Physics
GCSE Biology
Others English Help
ESL, IELTS, TOEFL
Programming
Animal Facts
Tutoring Services
What's New

 

Factor Theorem

In this lesson, we will look at the Factor Theorem

 

 

When f(x) is divided by xa, we get

f(x) = (xa)Q(x) + remainder

From the Remainder Theorem, we get

f(x) = (xa)Q(x) + f(a)

If f(a) = 0 then the remainder is 0 and

f(x) = (xa)Q(x)

We can then say that xa is a factor of f(x)

 

The Factor Theorem states that

xa is a factor of the polynomial f(x) if f(a) = 0

 

 

Example:

Determine whether x + 1 is a factor of the following polynomials.

a) 3x4 + x3x2 + 3x + 2

b) x6 + 2x(x – 1) – 4

Solution:

a) Let f(x) = 3x4 + x3x2 + 3x + 2

f(–1) = 3(–1)4 + (–1)3 – (–1)2 +3(–1) + 2

= 3(1) + (–1) – 1 – 3 + 2 = 0

Therefore, x + 1 is a factor of f(x)

 

b) Let g(x) = x6 + 2x(x – 1) – 4

g(–1) = (–1)6 + 2(–1)( –2) –4 = 1

Therefore, x + 1 is not a factor of g(x)

 

 

Videos

The Factor Theorem and The Remainder Theorem
What the theorems are and how they can be used to find the linear factorization of a polynomial.

 

 

This video demonstrates how to use the Factor Theorem to factor polynomials.

 

 

 

Custom Search

 

We welcome your feedback, comments and questions about this site - please submit your feedback via our Feedback page.

 

© Copyright 2005, 2009 - onlinemathlearning.com
Embedded content, if any, are copyrights of their respective owners.

Useful Links:
More Algebra Help at MathWorld
 

 

 

Custom Search