Remainder & Factor Theorem

 

 

 

Contents

 

Introduction

 

Remainder Theorem

 

Factor Theorem

 

Cubic Equations

 

Quiz 

 

 

Algebra Main Page

 

 

Solving Cubic Equation

By using factor theorem, together  with some intelligent guessing, we can factorise polynomials of higher degree.

In summary, to solve a cubic equation of form ax3 + bx2 + cx + d = 0,

1. Obtain one factor (x -a ) by trial and error.

2. Factorise ax3 + bx2 + cx + d = 0 as (x - a) (hx2 + kx + s) = 0

3. Solve the quadratic expression for other roots.

Example 1:

Factorise x3 - 2x2 - 5x + 6

  Let f(x) = x3 - 2x2 -5x +6

By inspection, f(x) if of the 3rd degree, we would expect it to have 3 linear factors at most so that 

f(x) = (x + a) (x + b) (x + c)

where a, b and c can be both positive and negative numbers. Also, by multiplying the last term of each factor, a x b x c numerically equals 6. Hence a, b and c are factors of 6: the possible factors of f(x) will then be (x ± 1), (x ± 2), (x ± 3) or (x ± 6). We must find the first one from these by trial.

Try (x - 1):

f(1) = 1 - 2 - 5 + 6 = 0 

We can now either do long division (a) or write it out as an identity (b).

(A) (B)
        x2- x - 6             
x - 1 ) x3 - 2x2- 5x + 6
          x3 - x2
              - x2 - 5x
              - x2 + x
                     - 6x - 6
                     - 6x - 6  
Let f(x) = (x - 1)(Ax2 + Bx + C)
A = 1                                 C = -6

let x = -1

(-1)3 - 2(-1)2 - 5(-1) + 6 = (-1 -1)(1(-1)2 -B - 6)
8 = -2(-5 - B)
B = -1
 
f(x) = (x - 1)(x2- x - 6)  

f(x) = (x - 1)(x - 3)(x + 2)

x = 1 or 3 or -2  

 
Example 2:
 
 Solve the equation 3x3 - 8x2 + 3x + 2 = 0 and hence find the value of y, such that 3e3y - 8e2y + 3ey + 2 = 0
 
Let f(x) = 3x3 - 8x2 + 3x + 2
         f (1) = 3 - 8 + 3 + 2 = 0
(x - 1) is a factor of f(x).
 
Factorising,
               (x - 1)(3x2 - 5x - 2) = 0
             (x - 1)(3x + 1)(x - 2) = 0
                x = 1 or -1/3 or 2
 
3e3y - 8e2y + 3e y + 2 = 0
3(ey)3 - 8(ey)2 + 3(ey) + 2 = 0
ey= 1 or 2 or -1/3 (n.a.)
 
y = 0 or ln 2

 
Example 3:
 
Given that f(x) = x3, - x2 - ax - b, where a and b are constants, has the factor x - 3 but has a remainder of 13x - 11 when divided by x + 4. Calculate the values of a and b. Hence (i) factorise f(x) completely. (ii) Solve the equation 27x3 - 3ax = ax2 + b.
 
f(3) = 0
33 - 32 - 3a - b = 0
3a + b = 18 --------(1)
 
f(-4) = (-4)3 - (-4)2 - a(-4) - b = 13(-4) - 11
4a - b = 17 --------(2)
 
sub (1) into (2): a = 5, b = 3
 
(i) Factorising.
    f(x) = (x - 3)(x2 + 2x + 1)
          = (x - 3)(x + 1)2
 
(ii) 27x3 - 9x2 - 3ax - b = 0
     (3x)3 - (3x)2 - a(3x) - b = 0
Let y = 3x
y3 - y2 - ay - b = 0
(y - 3)(y + 1)2 = 0        [from (i)]
y = 3, y = -1
x = 1, x = -1/3