Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 8 - Remainder And Factor Theorems

Share this:

Selina Textbook Solutions Chapter 8 - Remainder And Factor Theorems

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 8 - Remainder And Factor Theorems.

All Selina textbook questions of Chapter 8 - Remainder And Factor Theorems solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam. 

Read more
Exercise/Page

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 8 - Remainder And Factor Theorems Page/Excercise 8(A)

Solution 1

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

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.

 

 

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).

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.

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

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.

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

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

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.

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.

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

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

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.

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.

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.

Solution 16

Solution 17

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 8 - Remainder And Factor Theorems Page/Excercise 8(B)

Solution 1

(i) Let f(x) = x3 - 2x2 - 9x + 18

x - 2 = 0 x = 2

 

Remainder = f(2)

= (2)3 - 2(2)2 - 9(2) + 18

= 8 - 8 - 18 + 18

= 0

Hence, (x - 2) is a factor of f(x).

 

Now, we have:

x3 - 2x2 - 9x + 18 = (x - 2) (x2 - 9) = (x - 2) (x + 3) (x - 3)

 

(ii) Let f(x) = 2x3 + 5x2 - 28x - 15

x + 5 = 0 x = -5

 

Remainder = f(-5)

= 2(-5)3 + 5(-5)2 - 28(-5) - 15

= -250 + 125 + 140 - 15

= -265 + 265

= 0

Hence, (x + 5) is a factor of f(x).

 

Now, we have:

2x3 + 5x2 - 28x - 15 = (x + 5) (2x2 - 5x - 3)

= (x + 5) [2x2 - 6x + x - 3]

= (x + 5) [2x(x - 3) + 1(x - 3)]

= (x + 5) (2x + 1) (x - 3)

 

(iii) Let f(x) = 3x3 + 2x2 - 3x - 2

3x + 2 = 0

 

Hence, (3x + 2) is a factor of f(x).

 

Now, we have:

Solution 2

(i)

 

(ii) Let f(x) = 2x3 + x2 - 13x + 6

For x = 2,

f(x) = f(2) = 2(2)3 + (2)2 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0

Hence, (x - 2) is a factor of f(x).

 

 

(iii) f(x) = 3x3 + 2x2 - 23x - 30

For x = -2,

f(x) = f(-2) = 3(-2)3 + 2(-2)2 - 23(-2) - 30

= -24 + 8 + 46 - 30 = -54 + 54 = 0

Hence, (x + 2) is a factor of f(x).

 

 

(iv) f(x) = 4x3 + 7x2 - 36x - 63

For x = 3,

f(x) = f(3) = 4(3)3 + 7(3)2 - 36(3) - 63

= 108 + 63 - 108 - 63 = 0

Hence, (x + 3) is a factor of f(x).

 

 

(v) f(x) = x3 + x2 - 4x - 4

For x = -1,

f(x) = f(-1) = (-1)3 + (-1)2 - 4(-1) - 4

= -1 + 1 + 4 - 4 = 0

Hence, (x + 1) is a factor of f(x).

 

Solution 3

Let f(x) = 3x3 + 10x2 + x - 6

For x = -1,

f(x) = f(-1) = 3(-1)3 + 10(-1)2 + (-1) - 6 = -3 + 10 - 1 - 6 = 0

Hence, (x + 1) is a factor of f(x).

Now, 3x3 + 10x2 + x - 6 = 0

Solution 4

f (x) = 2x3 - 7x2 - 3x + 18

For x = 2,

f(x) = f(2) = 2(2)3 - 7(2)2 - 3(2) + 18

= 16 - 28 - 6 + 18 = 0

Hence, (x - 2) is a factor of f(x).

Solution 5

f(x) = x3 + 3x2 + ax + b

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

rightwards double arrow (2)3 + 3(2)2 + a(2) + b = 0

rightwards double arrow 8 + 12 + 2a + b = 0

rightwards double arrow 2a + b + 20 = 0 ...(i)

 

Since, (x + 1) is a factor of f(x), f(-1) = 0

rightwards double arrow (-1)3 + 3(-1)2 + a(-1) + b = 0

rightwards double arrow -1 + 3 - a + b = 0

rightwards double arrow -a + b + 2 = 0 ...(ii)

 

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

3a + 18 = 0

rightwards double arrow a = -6

 

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

b = a - 2 = -6 - 2 = -8

 

f(x) = x3 + 3x2 - 6x - 8

Now, for x = -1,

f(x) = f(-1) = (-1)3 + 3(-1)2 - 6(-1) - 8 = -1 + 3 + 6 - 8 = 0

Hence, (x + 1) is a factor of f(x).

 

Solution 6

Let f(x) = 4x3 - bx2 + x - c

 

It is given that when f(x) is divided by (x + 1), the remainder is 0.

f(-1) = 0

4(-1)3 - b(-1)2 + (-1) - c = 0

-4 - b - 1 - c = 0

b + c + 5 = 0 ...(i)

 

It is given that when f(x) is divided by (2x - 3), the remainder is 30.

 

Multiplying (i) by 4 and subtracting it from (ii), we get,

5b + 40 = 0

b = -8

 

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

c = -5 + 8 = 3

 

Therefore, f(x) = 4x3 + 8x2 + x - 3

 

Now, for x = -1, we get,

f(x) = f(-1) = 4(-1)3 + 8(-1)2 + (-1) - 3 = -4 + 8 - 1 - 3 = 0

 

Hence, (x + 1) is a factor of f(x).

 

Solution 7

f(x) = x2 + px + q

It is given that (x + a) is a factor of f(x).

g(x) = x2 + mx + n

It is given that (x + a) is a factor of g(x).

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

pa - q = ma - n

n - q = a(m - p)

Hence, proved.

Solution 8

Let f(x) = ax3 + 3x2 - 3

When f(x) is divided by (x - 4), remainder = f(4)

f(4) = a(4)3 + 3(4)2 - 3 = 64a + 45

Let g(x) = 2x3 - 5x + a

When g(x) is divided by (x - 4), remainder = g(4)

g(4) = 2(4)3 - 5(4) + a = a + 108

It is given that f(4) = g(4)

64a + 45 = a + 108

63a = 63

a = 1

Solution 9

Let f(x) = x3 - ax2 + x + 2

It is given that (x - a) is a factor of f(x).

Remainder = f(a) = 0

a3 - a3 + a + 2 = 0

a + 2 = 0

a = -2

Solution 10

Let the number to be subtracted from the given polynomial be k.

Let f(y) = 3y3 + y2 - 22y + 15 - k

It is given that f(y) is divisible by (y + 3).

Remainder = f(-3) = 0

3(-3)3 + (-3)2 - 22(-3) + 15 - k = 0

-81 + 9 + 66 + 15 - k = 0

9 - k = 0

k = 9

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 8 - Remainder And Factor Theorems Page/Excercise 8(C)

Solution 1

Let f(x) = x3 - 7x2 + 14x - 8

f(1) = (1)3 - 7(1)2 + 14(1) - 8 = 1 - 7 + 14 - 8 = 0

Hence, (x - 1) is a factor of f(x).

Solution 2

 

  

Solution 3

Let f(x) = x3 + 3x2 - mx + 4

According to the given information,

f(2) = m + 3

(2)3 + 3(2)2 - m(2) + 4 = m + 3

8 + 12 - 2m + 4 = m + 3

24 - 3 = m + 2m

3m = 21

m = 7

Solution 4

Let the required number be k.

Let f(x) = 3x3 - 8x2 + 4x - 3 - k

According to the given information,

f (-2) = 0

3(-2)3 - 8(-2)2 + 4(-2) - 3 - k = 0

-24 - 32 - 8 - 3 - k = 0

-67 - k = 0

k = -67

Thus, the required number is -67.

Solution 5

Let f(x) = x3 + (a + 1)x2 - (b - 2)x - 6

 

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

Remainder = f(-1) = 0

(-1)3 + (a + 1)(-1)2 - (b - 2) (-1) - 6 = 0

-1 + (a + 1) + (b - 2) - 6 = 0

a + b - 8 = 0 ...(i)

 

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

Remainder = f(2) = 0

(2)3 + (a + 1) (2)2 - (b - 2) (2) - 6 = 0

8 + 4a + 4 - 2b + 4 - 6 = 0

4a - 2b + 10 = 0

2a - b + 5 = 0 ...(ii)

 

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

3a - 3 = 0

a = 1

 

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

1 + b - 8 = 0

b = 7

 

f(x) = x3 + 2x2 - 5x - 6

Now, (x + 1) and (x - 2) are factors of f(x). Hence, (x + 1) (x - 2) = x2 - x - 2 is a factor of f(x).

 

 

f(x) = x3 + 2x2 - 5x - 6 = (x + 1) (x - 2) (x + 3)

Solution 6

Let f(x) = x2 + ax + b

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

Remainder = f(2) = 0

(2)2 + a(2) + b = 0

4 + 2a + b = 0

2a + b = -4 ...(i)

 

It is given that:

a + b = 1 ...(ii)

 

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

a = -5

 

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

b = 1 - (-5) = 6

Solution 7

Let f(x) = x3 + 6x2 + 11x + 6

For x = -1

f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6

= -1 + 6 - 11 + 6 = 12 - 12 = 0

Hence, (x + 1) is a factor of f(x).

Solution 8

Let f(x) = mx3 + 2x2 - 3

g(x) = x2 - mx + 4

It is given that f(x) and g(x) leave the same remainder when divided by (x - 2). Therefore, we have:

f (2) = g (2)

m(2)3 + 2(2)2 - 3 = (2)2 - m(2) + 4

8m + 8 - 3 = 4 - 2m + 4

10m = 3

m =

Solution 9

Let f(x) = px3 + 4x2 - 3x + q

It is given that f(x) is completely divisible by (x2 - 1) = (x + 1)(x - 1).

 

Therefore, f(1) = 0 and f(-1) = 0

f(1) = p(1)3 + 4(1)2 - 3(1) + q = 0

p + q + 1 = 0 ...(i)

 

f(-1) = p(-1)3 + 4(-1)2 - 3(-1) + q = 0

-p + q + 7 = 0 ...(ii)

 

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

2q + 8 = 0

q = -4

 

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

p = -q - 1 = 4 - 1 = 3

 

f(x) = 3x3 + 4x2 - 3x - 4

 

Given that f(x) is completely divisible by (x2 - 1).

 

Solution 10

Let the required number be k.

Let f(x) = x2 + x + 3 + k

It is given that f(x) is divisible by (x + 3).

Remainder = 0

f (-3) = 0

(-3)2 + (-3) + 3 + k = 0

9 - 3 + 3 + k = 0

9 + k = 0

k = -9

Thus, the required number is -9.

Solution 11

It is given that when the polynomial x3 + 2x2 - 5ax - 7 is divided by (x - 1), the remainder is A.

(1)3 + 2(1)2 - 5a(1) - 7 = A

1 + 2 - 5a - 7 = A

- 5a - 4 = A ...(i)

 

It is also given that when the polynomial x3 + ax2 - 12x + 16 is divided by (x + 2), the remainder is B.

x3 + ax2 - 12x + 16 = B

(-2)3 + a(-2)2 - 12(-2) + 16 = B

-8 + 4a + 24 + 16 = B

4a + 32 = B ...(ii)

 

It is also given that 2A + B = 0

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

2(-5a - 4) + 4a + 32 = 0

-10a - 8 + 4a + 32 = 0

-6a + 24 = 0

6a = 24

a = 4

Solution 12

Let f(x) = (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15

It is given that (3x + 5) is a factor of f(x).

f(x) = (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15

= 3x3 + 5x2 - 9x - 15

Solution 13

If (x - 3) divides f(x) = x3 - px2 + x + 6, then,

Remainder = f(3) = 33 - p(3)2 + 3 + 6 = 36 - 9p

If (x - 3) divides g(x) = 2x3 - x2 - (p + 3) x - 6, then

Remainder = g(3) = 2(3)3 - (3)2 - (p + 3) (3) - 6 = 30 - 3p

Now, f(3) = g(3)

36 - 9p = 30 - 3p

-6p = -6

p = 1

Solution 14

f(x) = 2x3 + x2 - 13x + 6

Factors of constant term 6 are 1, 2, 3, 6.

Putting x = 2, we have:

f(2) = 2(2)3 + 22 - 13 (2) + 6 = 16 + 4 - 26 + 6 = 0

Hence (x - 2) is a factor of f(x).

Solution 15

Let f(x) = 2x3 + 3x2 - kx + 5 

Using Remainder Theorem, we have

f(2) = 7

2(2)3 + 3(2)2 - k(2) + 5 = 7

16 + 12 - 2k + 5 = 7

33 - 2k = 7

2k = 26

k = 13

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 8 - Remainder And Factor Theorems  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well. 

Text Book Solutions

ICSE X - Mathematics

This content is available for subscribed users only.

OR

Call us

1800-212-7858 (Toll Free) to speak to our academic expert.
OR

Let us get in touch with you

Chat with us on WhatsApp