REMAINDER THEOREM AND FACTOR THEOREM

In this section , we shall study a simple and an elegant method of finding the remainder.

In the case of divisibility of a polynomial by a linear polynomial we use a well known theorem called Remainder Theorem.

Example 1 :

Using Remainder Theorem, find the remainder when

f(x) = x³ + 3x² + 3x + 1

is divided by (x + 1).

Solution : 

Here, the divisor is (x + 1). 

Equate the divisor to zero. 

x + 1 = 0

Solve for x. 

x = -1

To find the remainder, substitute -1 for x into the function f(x). 

f(-1) = (-1)³ + 3(-1)² + 3(-1) + 1

f(-1) = -1 + 3(1) - 3 + 1

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

f(-1) = 0

So, the remainder is 0.

Example 2 :

Using Remainder Theorem, find the remainder when

f(x) = x³ - 3x + 1

is divided by (2 - 3x).

Solution : 

Here, the divisor is (2 - 3x). 

Equate the divisor to zero. 

2 - 3x = 0

Solve for x. 

-3x = -2

x = ⅔

To find the remainder, substitute 2/3 for x into the function f(x). 

f(⅔) = (⅔)³ - 3(⅔) + 1

= ⁸⁄₂₇ - 2 + 1

= ⁸⁄₂₇ - 1

= ⁸⁄₂₇ - ²⁷⁄₂₇

= ⁽⁸ ⁻ ²⁷⁾⁄₂₇

= -¹⁹⁄₂₇

So, the remainder is -¹⁹⁄₂₇.

Example 3 :

For what value of k is the polynomial

2x⁴ + 3x³ + 2kx² + 3x + 6

is divisible by (x + 2).

Solution : 

Let f(x) = 2x⁴ + 3x³ + 2kx² + 3x + 6.

Here, the divisor is (x + 2). 

Equate the divisor to zero. 

x + 2 = 0

Solve for x. 

x = -2

To find the remainder, substitute -2 for x into the function f(x). 

f(-2) = 2(-2)⁴ + 3(-2)³ + 2k(-2)² + 3(-2) + 6

f(-2) = 2(16) + 3(-8) + 2k(4) - 6 + 6

f(-2) = 32 - 24 + 8k - 6 + 6

f(-2) = 8 + 8k

So, the remainder is (8 + 8k).

If f(x) is exactly divisible by (x + 2), then the remainder must be zero.

Then, 

8 + 8k = 0

Solve for k.

8k = -8

k = -1

Therefore, f(x) is exactly divisible by (x + 2) when

k = –1

Example 1 :

Using Factor Theorem, show that (x + 2) is a factor of 

x³ - 4x² - 2x + 20

Solution : 

Let f(x) = x³ - 4x² - 2x + 20.

Equate the factor (x + 2) to zero.

x + 2 = 0

Solve for x. 

x = -2

By Factor Theorem,

(x + 2) is factor of f(x), if f(-2)  =  0

Then, 

f(-2) = (-2)³ - 4(-2)² - 2(-2) + 20

f(-2) = -8 - 4(4) + 4 + 20

f(-2) = -8 - 16 + 4 + 20

f(-2) = 0

Therefore, (x + 2) is a factor of x³ - 4x² - 2x + 20. 

Example 2 :

Is (3x - 2) a factor of 3x³ + x² - 20x + 12 ?

Solution : 

Let f(x)  =  3x³ + x² - 20x + 12.

Equate the factor (3x + 2) to zero.

3x - 2 = 0

Solve for x. 

3x = 2

x = ⅔

By Factor Theorem, (3x - 2) is factor of f(x), if

f(⅔) = 0

Then,

f(⅔) = 3(⅔)³ + (⅔)² - 20(⅔) + 12

= 3(⁸⁄₂₇) + ⁴⁄₉ - ⁴⁰⁄₃ + 12

= ⁸⁄₉ + ⁴⁄₉ - ⁴⁰⁄₃ + 12

= ⁸⁄₉ + ⁴⁄₉ - ¹²⁰⁄₉ + ¹⁰⁸⁄₉

= ⁽⁸ ⁺ ⁴ ⁻ ¹²⁰ ⁺ ¹⁰⁸⁾⁄₉

= ⁽¹²⁰ ⁻ ¹²⁰⁾⁄₉

= 0

Therefore, (3x - 2) is a factor of 3x³ + x² - 20x + 12. 

Example 3 :

Find the value of m, if (x - 2) is a factor of the polynomial

2x³ - 6x² + mx + 4

Solution : 

Let f(x) = 2x³ - 6x² + mx + 4.

Equate the factor (x - 2) to zero.

x - 2 = 0

Solve for x.

x = 2

By Factor Theorem,

(x - 2) is factor of f(x), if f(2) = 0

Then,

f(2) = 0

2(2)³ - 6(2)² + m(2) + 4 = 0

f(2)  =  2(8) - 6(4) + 2m + 4 = 0

f(2)  =  16 - 24 + 2m + 4 = 0

f(2)  =  2m - 4 = 0

2m = 4

m = 2

Therefore (x - 2) is a factor of f(x), when

m = 2 

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