Factor theorem is the extension of remainder theorem. We call an integer to be a factor of another integer, if it divides the later one exactly or leaves the remainder 0. The same procedure hold good for an algebraic expression also.

We know that a polynomial can be divided by another polynomial whose degree is less than that of the dividend. According to the remainder theorem, when a polynomial f (x) is divided by a linear factor (x - a), then the remainder obtained after dividing f(x) by (x - a) is f (a). Let us extend this to define the factor theorem.

In this section we shall also see the proof of factor theorem and some of the interesting examples.

A polynomial of any degree in f (x) is said to be exactly divisible by the linear factor (x - a), if the remainder f (a) = 0.

Example: Verify if (x - 2) is a factor of the polynomial x2 - 7x + 10.
Solution:
let f (x) = x2 - 7x + 10
Substituting x = 2, we get, f (2) = 22 - 7 (2) + 10 = 4 - 14 + 10 = 14 - 14 = 0
Since f (2) = 0, we see that (x - 2) is the factor.
Verification: We can divide the given polynomial by (x- 2) and check if it is exactly divisible by (x - 2)

Method 1: Factorization:

As the numerator x2 - 7x + 10 is factorisable, we can split the middle term and factorize.

x2 - 7x + 10 = x2 - 5x - 2 x + 10
= x (x - 5) - 2 (x - 5)
= (x - 5) (x - 2)

Therefore, $\frac{x^{2}-7x+10}{(x-2)}$ = $\frac{(x-5)(x-2)}{(x-2)}$

= (x - 5)
We get the quotient as (x - 5) and the Remainder 0.

Method 2: By long division method:


___x_-_5_______
(x - 2) | x2 - 7x + 10
x2 - 2 x
- 5 x + 10
- 5x + 10
0
We can see that the Quotient is (x - 5) and the Remainder is 0.

Method 3: By Synthetic Division

2 | 1 -7 10
| 2 -10
1 -5 0 ---> Remainder

Therefore, we see that the Quotient is (x - 5) and the Remainder is 0.

According to Factor Theorem, a polynomial f ( x ) is divisible by ( x- a ) if and only if the remainder f ( a ) is zero.

Case 1: Let the polynomial f (x ) be divisible by ( x - a )
To prove : f ( a ) = 0
Proof : Since f (x ) is divisible by ( x - a ),
we have, f ( x ) = q ( x ) . ( x - a )
Substituting x = a, we get, f (a ) = q ( a ). ( a - a )
f (a ) = q ( a ) . 0
f (a ) = 0

Case 2: Let f ( a ) = 0
To prove : ( x - a ) is a factor of f ( x ).
Proof : According to remainder theorem when f (x ) is divided by ( x - a ), the remainder = f ( a ),
Therefore, f ( x ) = q ( x ) . ( x - a ) + f (a ) where f ( a ) is the remainder.
Since we are given that f ( a ) = 0, we get,
f ( x ) = q ( x ) ( x - a ) + 0
f ( x ) = q ( x ) . ( x - a )
=> ( x - a ) is the factor of f ( x ).

Solved Examples

Question 1:

Using factor theorem show that ( x + 1 ) is a factor of 2 x3 + x2 - 2x - 1.


Solution:
 
We have f ( x ) = 2 x3 + x2 - 2x - 1
According to factor theorem, if ( x - a ) is a factor of f ( x ), then f (a ) = 0.
Therefore ( x + 1 ) = ( x - a )
     => ( x - ( -1 ) ) = ( x - a )
     =>              a = - 1
Substituting x = - 1, in the polynomial f ( x ), we get,
                 f ( -1 ) = 2 ( -1 )3 + ( -1 )2 - 2 ( -1 ) -1
                          = 2 ( -1 ) + 1 + 2 - 1
                          = -2 + 2
                          = 0
Since the remainder f ( -1 ) = 0, we get ( x + 1 ) is a factor of f ( x ).

 

Question 2:

Using factor theorem show that ( x - 2 ) is a factor of the polynomial 2 x - x2 -  5x - 2. Also find the other roots.


Solution:
 
We have f ( x ) = 2 x - x2 -  5x - 2
According to factor theorem,( x - 2 ) be a factor of given polynomial if f ( 2 ) is equal to zero.
Therefore, substituting x = 2 in f ( x ), we get,
                f ( 2 ) = 2 ( 2 )3 - ( 2 )2 - 5 ( 2 ) - 2
                         = 2 ( 8 ) - 4 - 10 - 2
                         = 16 - 16
                         = 0
Therefore, ( x - 2 ) is the factor of f ( x ).

Dividing f ( x ) by ( x - 2), using synthetic division, we get,

                2   |    2             -1             -5          -2
                     |    0              4              6           2  
                         2              3               1       |  0  ------> Remainder
The quotient q ( x ) = 2 x2 + 3x + 1
                            = 2 x2 + 2x + x + 1
                            = 2x ( x + 1 ) + 1 ( x + 1 )
                            = ( x + 1 ) ( 2x + 1 )
Therefore, the complete factors of f (x ) are, ( x - 2 ), ( x + 1 ) and ( 2x + 1 ).
 

Question 3:

Show that ( x - 2 ) is a factor of the polynomial, 2 x 3 + 5 x- 11x - 14


Solution:
 
Let f ( x ) = 2 x 3 + 5 x- 11x - 14
To prove that ( x - 2 ) is a factor of f ( x ), we need to prove that f ( 2 ) = 0,
substituting x = 2, in the above polynomial, we get,
                     f ( 2 ) = 2 ( 2 )+ 5 ( 2 )- 11 ( 2 ) - 14
                             = 16 + 20 - 22 - 14
                             = 36 - 36
                             = 0
Since the factor theorem is satisfied, we prove that ( x - 2 ) is a factor of f ( x ).