REMAINDER THEOREM AND FACTOR THEOREM

Remainder Theorem and Factor Theorem :

In this section, you will learn Remainder Theorem and Factor Theorem. 

Remainder 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)

Remainder Theorem - Practice Problems

Problem 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.

Problem 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  =  2/3

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

f(2/3)  =  (2/3)³ - 2/3 + 1

f(2/3)  =  8/27 - 18/27 + 27/27

f(2/3)  =  (8 - 18 + 27) / 27

f(2/3)  =  17 / 27

So, the remainder is 17/27.

Problem 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.

Factor Theorem - Practice Problems 

Problem 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. 

Problem 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  =  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 3x³ + x² - 20x + 12. 

Problem 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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