Remainder & Factor Theorem

 

 

 

Contents

 

Introduction

 

Remainder Theorem

 

Factor Theorem

 

Cubic Equations

 

Quiz 

 

 

Algebra Main Page

 

 

Factor Theorem

Factor Theorem is a special case of Remainder Theorem

f(x) = (x - k) Q(x) + R

f(x) = (x - k) Q(x) + f(k)

when f(k) = 0,

f(x) = (x - k) Q(x)

therefore, f(x) is exactly divisible by x - k.

Example 1

Given that x2 + x - 6 is a factor of 2x4 + x3 - ax + bx + a + b - 1, find the value of a and b.

  x2 + x - 6 = (x + 3)(x + 2)

      Let f(x) =  2x4 + x3 - ax2 + bx + a + b - 1
f(-3) = 2(-3)4 + (-3)3 - a(-3)2 - 3b + a + b - 1 = 0
           134 - 8a - 2b = 0
           4a + b = 67 --------(1)

f(2) = 2(2)4 + 23 - a(2)2 + 2b + a + b - 1 = 0

         39 - 3a + 3b = 0
         a - b = 13 --------(2)

(1) + (2) : 5a = 80

                a = 16
when a = 16, b = 3

Example 2

 
 Use the factor theorem to find the value of k for which (a + 2b) where a does not = 0 and b does not = 0, is a factor of a4 + 32b4 + a3b(k + 3)   
 
Let f(a) = a4 + 32b4 + a3b(k + 3)
       f(-2b) = (-2b)4 + 32b4 + (-2b)3b(k + 3) = 0
                    48b4 - 8b4(k+ 3) = 0
                    8b4[6 - (k + 3)] = 0
                    8b4(3 - k) = 0
 
Since b does not = 0, 3 - k = 0
                           k = 3

 
Example 3
 
Determine the value of k for which x + 2 is a factor of (x + 1)7 + (2x + k)3.
 
Let f(x) = (x + 1) + (2x + k)
         f(-2) = (-2 + 1) + (-4 + k) = 0
                   (k - 4) - 1 = 0
                   (k - 4) = 1
                   k - 4 = 1
                   k = 5