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.

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) = x^{3} + 3x^{2} + 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) = x^{3} - 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 = 2/3

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

f(2/3) = (2/3)^{3} - 3(2/3) + 1

f(2/3) = 8/27 - 2 + 1

f(2/3) = 8/27 - 1

f(2/3) = 8/27 - 27/27

f(2/3) = (8 - 27)/27

f(2/3) = -19/27

So, the remainder is -19/27.

**Example 3 :**

For what value of k is the polynomial

2x^{4} + 3x^{3} + 2kx^{2} + 3x + 6

is divisible by (x + 2).

**Solution : **

Let

f(x) = 2x^{4} + 3x^{3} + 2kx^{2} + 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.

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

x^{3} - 4x^{2} - 2x + 20

**Solution : **

Let

f(x) = x^{3} - 4x^{2} - 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 x^{3} - 4x^{2} - 2x + 20.

**Example 2 :**

Is (3x - 2) a factor of 3x^{3} + x^{2} - 20x + 12 ?

**Solution : **

Let

f(x) = 3x^{3} + x^{2} - 20x + 12

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

3x - 2 = 0

Solve for x.

3x = 2

x = 2/3

By Factor Theorem,

(3x - 2) is factor of f(x), if f(2/3) = 0

Then,

f(2/3) = 3(2/3)^{3} + (2/3)^{2} - 20(2/3) + 12

f(2/3) = 3(8/27) + 4/9 - 40/3 + 12

f(2/3) = 8/9 + 4/9 - 40/3 + 12

f(2/3) = 8/9 + 4/9 - 120/9 + 108/9

f(2/3) = (8 + 4 - 120 + 108) / 9

f(2/3) = (120 - 120) / 9

f(2/3) = 0

Therefore, (3x - 2) is a factor of 3x^{3} + x^{2} - 20x + 12.

**Example 3 :**

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

2x^{3} - 6x^{2} + mx + 4

**Solution : **

Let

f(x) = 2x^{3} - 6x^{2} + 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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