# 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.

## Remainder Theorem

If a polynomial p(x) of degree greater than or equal to one is divided by a linear polynomial (x–a) then the remainder is p(a), where a is any real number.

Significance of Remainder theorem :

It enables us to find the remainder without actually following the cumbersome process of long division.

Note :

(i) If p(x) is divided by (x+a), then the remainder is

p(–a)

(ii) If p(x) is divided by (ax–b), then the remainder is

p(b⁄ₐ)

(iii) If p(x) is divided by (ax+b), then the remainder is

p(-b⁄ₐ)

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 =

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

f() = ()3 - 3() + 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

## Factor Theorem

If p(x) is a polynomial of degree n ≥ 1 and a is any real number then

(i) p(a)  =  0 implies (x - a) is a factor of p(x).

(ii) (x - a) is a factor of p(x) implies p(a)  =  0.

Note :

(i) (x - a) is a factor of p(x), if p(a)  =  0.

(ii) (x + a) is a factor of p(x), if p(-a)  =  0.

(iii) (ax + b) is a factor of p(x), if p(-b⁄ₐ)  =  0.

(iv) (x - a)(x - b) is a factor of p(x), if

p(a)  =  0  and  p(b)  =  0

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 =

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

f() = 0

Then,

f() = 3()3 + ()2 - 20() + 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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