Chapter 8 - Remainder And Factor Theorems Exercise Ex. 8(A)

Question 1

Find, in each case, the remainder when:

Solution 1

By remainder theorem we know that when a polynomial f (x) is divided by x - a, then the remainder is f(a).

Question 2

Show that:

Solution 2

(x - a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x - a), is 0, i.e., if f(a) = 0.

 

 

Question 3

Use the Remainder Theorem to find which of the following is a factor of 2x³ + 3x² - 5x - 6.

(i) x + 1

(ii) 2x - 1

(iii) x + 2

Solution 3

By remainder theorem we know that when a polynomial f (x) is divided by x - a, then the remainder is f(a).

Let f(x) = 2x³ + 3x² - 5x - 6

 

(i) f (-1) = 2(-1)³ + 3(-1)² - 5(-1) - 6 = -2 + 3 + 5 - 6 = 0

Thus, (x + 1) is a factor of the polynomial f(x).

 

(ii)

Thus, (2x - 1) is not a factor of the polynomial f(x).

 

(iii) f (-2) = 2(-2)³ + 3(-2)² - 5(-2) - 6 = -16 + 12 + 10 - 6 = 0

Thus, (x + 2) is a factor of the polynomial f(x).

Question 4

(i) If 2x + 1 is a factor of 2x² + ax - 3, find the value of a.

(ii) Find the value of k, if 3x - 4 is a factor of expression 3x² + 2x - k.

Solution 4

(i) 2x + 1 is a factor of f(x) = 2x² + ax - 3.

(ii) 3x - 4 is a factor of g(x) = 3x² + 2x - k.

Question 5

Find the values of constants a and b when x - 2 and x + 3 both are the factors of expression x³ + ax² + bx - 12.

Solution 5

Let f(x) = x³ + ax² + bx - 12

x - 2 = 0  x = 2

x - 2 is a factor of f(x). So, remainder = 0

x + 3 = 0  x = -3

x + 3 is a factor of f(x). So, remainder = 0

Adding (1) and (2), we get,

5a - 15 = 0

 a = 3

Putting the value of a in (1), we get,

6 + b - 2 = 0

 b = -4

Question 6

find the value of k, if 2x + 1 is a factor of (3k + 2)x³ + (k - 1). 

Solution 6

Let f(x) = (3k + 2)x³ + (k - 1)

 

2x + 1 = 0 

Since, 2x + 1 is a factor of f(x), remainder is 0.

Question 7

Find the value of a, if x - 2 is a factor of 2x⁵ - 6x⁴ - 2ax³ + 6ax² + 4ax + 8.

Solution 7

f(x) = 2x⁵ - 6x⁴ - 2ax³ + 6ax² + 4ax + 8

x - 2 = 0  x = 2

Since, x - 2 is a factor of f(x), remainder = 0.

2(2)⁵ - 6(2)⁴ - 2a(2)³ + 6a(2)² + 4a(2) + 8 = 0

64 - 96 - 16a + 24a + 8a + 8 = 0

-24 + 16a = 0

16a = 24

a = 1.5

Question 8

Find the values of m and n so that x - 1 and x + 2 both are factors of 

x³ + (3m + 1) x² + nx - 18.

Solution 8

Let f(x) = x³ + (3m + 1) x² + nx - 18

x - 1 = 0  x = 1

x - 1 is a factor of f(x). So, remainder = 0

x + 2 = 0  x = -2

x + 2 is a factor of f(x). So, remainder = 0

Adding (1) and (2), we get,

9m - 27 = 0

m = 3

Putting the value of m in (1), we get,

3(3) + n - 16 =0

9 + n - 16 = 0

n = 7

Question 9

When x³ + 2x² - kx + 4 is divided by x - 2, the remainder is k. Find the value of constant k.

Solution 9

Let f(x) = x³ + 2x² - kx + 4

x - 2 = 0  x = 2

On dividing f(x) by x - 2, it leaves a remainder k.

Question 10

Find the value of a, if the division of ax³ + 9x² + 4x - 10 by x + 3 leaves a remainder 5.

Solution 10

Let f(x) = ax³ + 9x² + 4x - 10

x + 3 = 0  x = -3

On dividing f(x) by x + 3, it leaves a remainder 5.

Question 11

If x³ + ax² + bx + 6 has x - 2 as a factor and leaves a remainder 3 when divided by x - 3, find the values of a and b.

Solution 11

Let f(x) = x³ + ax² + bx + 6

x - 2 = 0  x = 2

Since, x - 2 is a factor, remainder = 0

x - 3 = 0  x = 3

On dividing f(x) by x - 3, it leaves a remainder 3.

Subtracting (i) from (ii), we get,

a + 3 = 0

a = -3

Substituting the value of a in (i), we get,

-6 + b + 7 = 0

b = -1

Question 12

The expression 2x³ + ax² + bx - 2 leaves remainder 7 and 0 when divided by 2x - 3 and x + 2 respectively. Calculate the values of a and b.

Solution 12

Let f(x) = 2x³ + ax² + bx - 2

2x - 3 = 0  x = 

On dividing f(x) by 2x - 3, it leaves a remainder 7.

x + 2 = 0  x = -2

On dividing f(x) by x + 2, it leaves a remainder 0.

Adding (i) and (ii), we get,

7a - 21 = 0

a = 3

Substituting the value of a in (i), we get,

9 + 2b - 3 = 0

2b = -6

b = -3

Question 13

What number should be added to 3x³ - 5x² + 6x so that when resulting polynomial is divided by x - 3, the remainder is 8?

Solution 13

Let the number k be added and the resulting polynomial be f(x).

So, f(x) = 3x³ - 5x² + 6x + k

It is given that when f(x) is divided by (x - 3), the remainder is 8.

Thus, the required number is -46.

Question 14

What number should be subtracted from x³ + 3x² - 8x + 14 so that on dividing it with x - 2, the remainder is 10.

Solution 14

Let the number to be subtracted be k and the resulting polynomial be f(x).

So, f(x) = x³ + 3x² - 8x + 14 - k

It is given that when f(x) is divided by (x - 2), the remainder is 10.

Thus, the required number is 8.

Question 15

The polynomials 2x³ - 7x² + ax - 6 and x³ - 8x² + (2a + 1)x - 16 leaves the same remainder when divided by x - 2. Find the value of 'a'.

Solution 15

Let f(x) = 2x³ - 7x² + ax - 6

 x - 2 = 0  x = 2

 When f(x) is divided by (x - 2), remainder = f(2)

 Let g(x) = x³ - 8x² + (2a + 1)x - 16

 When g(x) is divided by (x - 2), remainder = g(2)

By the given condition, we have:

f(2) = g(2)

2a - 18 = 4a - 38

4a - 2a = 38 - 18

2a = 20

a = 10

 Thus, the value of a is 10.

Question 16

If (x - 2) is a factor of the expression 2x³ + ax² + bx - 14 and when the expression is divided by (x - 3), it leaves a remainder 52, find the values of a and b

Solution 16

Question 17

Find 'a' if the two polynomials ax³ + 3x² - 9 and 2x³ + 4x + a, leave the same remainder when divided by x + 3.

Solution 17