# Remainder And Factor Theorems Concise Maths Solution Chapter-8

## Selina Publications Chapter-8 ICSE Board Mathematics Class 10

**Remainder And Factor Theorems** Concise Maths Solution Chapter-8 for ICSE Board Class 10th Concise Solution **Remainder And Factor Theorems **Concise Maths Solution Chapter-8 for ICSE Maths Class 10 is available here. All Solution of Concise of Chapter 8 **Remainder And Factor Theorems** has been solved according instruction given by council. This is the Solution of Chapter-8 **Remainder And Factor Theorems** for ICSE Class 10th .ICSE Maths text book of Concise is In series of famous ICSE writer in maths. Concise is most famous among students. With the help of Concise Solution student can achieve their goal in 2020 exam of council.

**Remainder And Factor Theorems Concise Maths Solution Chapter-8**

The Solution of Concise Mathematics Chapter8 **Remainder And Factor Theorems** for ICSE Class 10 have been solved by experience teachers from across the globe to help students of class 10th ICSE board exams conducted by the ICSE (Indian Council of Secondary Education) board papering in 2020. Therefore the ICSE Class 10th Maths Solution of Concise solve problems of exercise related to various topics which are prescribed in most ICSE Maths textbooks.

**Chapter-8 Remainder And Factor Theorems Concise Maths Solution for ICSE Board Class 10th— Select Topics**

**Exe – 8 (A) , Exe -8 (B) , Exe – 8 (C) ,**

**How to Solve Concise Maths Selina Publications Chapter-8 Remainder And Factor Theorems**

Note :- Before viewing Solutions of Chapter -8**Remainder And Factor Theorems **of Concise Maths read the Chapter Carefully then solve all example of your text book. The Chapter- 8 **Remainder And Factor Theorems** is main Chapter in ICSE board

**Exercise – 8(A), Remainder And Factor Theorems Concise Maths Solution **

**Question 1**

Find, in each case, the remainder when:

**Answer 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:

#### Answer 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 2x^{3} + 3x^{2} – 5x – 6.

(i) x + 1

(ii) 2x – 1

(iii) x + 2

#### Answer 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) = 2x^{3} + 3x^{2} – 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 2x^{2} + ax – 3, find the value of a.

(ii) Find the value of k, if 3x – 4 is a factor of expression 3x^{2} + 2x – k.

#### Answer 4

(i) 2x + 1 is a factor of f(x) = 2x^{2} + ax – 3.

(ii) 3x – 4 is a factor of g(x) = 3x^{2} + 2x – k.

#### Question 5

Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x^{3} + ax^{2} + bx – 12.

#### Answer 5

Let f(x) = x^{3} + ax^{2} + 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)x^{3} + (k – 1).

#### Answer 6

Let f(x) = (3k + 2)x^{3} + (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 2x^{5} – 6x^{4} – 2ax^{3} + 6ax^{2} + 4ax + 8.

#### Answer 7

f(x) = 2x^{5} – 6x^{4} – 2ax^{3} + 6ax^{2} + 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 x^{3} + (3m + 1) x^{2} + nx – 18.

#### Answer 8

Let f(x) = x^{3} + (3m + 1) x^{2} + 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 x^{3} + 2x^{2} – kx + 4 is divided by x – 2, the remainder is k. Find the value of constant k.

#### Answer 9

Let f(x) = x^{3} + 2x^{2} – 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 ax^{3} + 9x^{2} + 4x – 10 by x + 3 leaves a remainder 5.

#### Answer 10

Let f(x) = ax^{3} + 9x^{2} + 4x – 10

x + 3 = 0 x = -3

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

#### Question 11

If x^{3} + ax^{2} + 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.

#### Answer 11

Let f(x) = x^{3} + ax^{2} + 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 2x^{3} + ax^{2} + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.

#### Answer 12

Let f(x) = 2x^{3} + ax^{2} + 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 3x^{3} – 5x^{2} + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8?

#### Answer 13

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

So, f(x) = 3x^{3} – 5x^{2} + 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 x^{3} + 3x^{2} – 8x + 14 so that on dividing it with x – 2, the remainder is 10.

#### Answer 14

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

So, f(x) = x^{3} + 3x^{2} – 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 2x^{3} – 7x^{2} + ax – 6 and x^{3} – 8x^{2} + (2a + 1)x – 16 leaves the same remainder when divided by x – 2. Find the value of ‘a’.

#### Answer 15

Let f(x) = 2x^{3} – 7x^{2} + ax – 6

x – 2 = 0 x = 2

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

Let g(x) = x^{3} – 8x^{2} + (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 2x^{3} + ax^{2} + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b

#### Answer 16

#### Question 17

Find ‘a’ if the two polynomials ax^{3} + 3x^{2} – 9 and 2x^{3} + 4x + a, leave the same remainder when divided by x + 3.

#### Answer 17

**Concise Maths Solution Exercise – 8(B) Remainder And Factor Theorems**

#### Question 1

Using the Factor Theorem, show that:

(i) (x – 2) is a factor of x^{3} – 2x^{2} – 9x + 18. Hence, factorise the expression x^{3} – 2x^{2} – 9x + 18 completely.

(ii) (x + 5) is a factor of 2x^{3} + 5x^{2} – 28x – 15. Hence, factorise the expression 2x^{3} + 5x^{2} – 28x – 15 completely.

(iii) (3x + 2) is a factor of 3x^{3} + 2x^{2} – 3x – 2. Hence, factorise the expression 3x^{3} + 2x^{2} – 3x – 2 completely.

#### Answer 1

(i) Let f(x) = x^{3} – 2x^{2} – 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:

x^{3} – 2x^{2} – 9x + 18 = (x – 2) (x^{2} – 9) = (x – 2) (x + 3) (x – 3)

(ii) Let f(x) = 2x^{3} + 5x^{2} – 28x – 15

x + 5 = 0 ⇒ x = -5

Remainder = f(-5)

= 2(-5)^{3} + 5(-5)^{2} – 28(-5) – 15

and = -250 + 125 + 140 – 15

so = -265 + 265

Therefore = 0

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

Now, we have:

2x^{3} + 5x^{2} – 28x – 15 = (x + 5) (2x^{2} – 5x – 3)

= (x + 5) [2x^{2} – 6x + x – 3]

and = (x + 5) [2x(x – 3) + 1(x – 3)]

then = (x + 5) (2x + 1) (x – 3)

(iii) Let f(x) = 3x^{3} + 2x^{2} – 3x – 2

3x + 2 = 0

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

Now, we have:

#### Question 2

Using the Remainder Theorem, factorise each of the following completely.

(i) 3x^{3 }+ 2x^{2} − 19x + 6

(ii) 2x^{3} + x^{2} – 13x + 6

(iii) 3x^{3} + 2x^{2} – 23x – 30

(iv) 4x^{3} + 7x^{2} – 36x – 63

(v) x^{3} + x^{2} – 4x – 4

#### Answer 2

(i)

(ii) Let f(x) = 2x^{3} + x^{2} – 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) = 3x^{3} + 2x^{2} – 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) = 4x^{3} + 7x^{2} – 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) = x^{3} + x^{2} – 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).

#### Question 3

Using the Remainder Theorem, factorise the expression 3x^{3} + 10x^{2} + x – 6. Hence, solve the equation 3x^{3} + 10x^{2} + x – 6 = 0.

#### Answer 3

Let f(x) = 3x^{3} + 10x^{2} + 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, 3x^{3} + 10x^{2} + x – 6 = 0

#### Question 4

Factorise the expression f (x) = 2x^{3} – 7x^{2} – 3x + 18. Hence, find all possible values of x for which f(x) = 0.

#### Answer 4

f (x) = 2x^{3} – 7x^{2} – 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).

#### Question 5

Given that x – 2 and x + 1 are factors of f(x) = x^{3} + 3x^{2} + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

#### Answer 5

f(x) = x^{3} + 3x^{2} + ax + b

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

(2)^{3} + 3(2)^{2} + a(2) + b = 0

8 + 12 + 2a + b = 0

2a + b + 20 = 0 …(i)

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

(-1)^{3} + 3(-1)^{2} + a(-1) + b = 0

-1 + 3 – a + b = 0

-a + b + 2 = 0 …(ii)

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

3a + 18 = 0

a = -6

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

b = a – 2 = -6 – 2 = -8

f(x) = x^{3} + 3x^{2} – 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).

#### Question 6

The expression 4x^{3} – bx^{2} + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.

#### Answer 6

Let f(x) = 4x^{3} – bx^{2} + 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) = 4x^{3} + 8x^{2} + 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).

#### Question 7

If x + a is a common factor of expressions f(x) = x^{2} + px + q and g(x) = x^{2} + mx + n; show that:

#### Answer 7

f(x) = x^{2} + px + q

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

g(x) = x^{2} + 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.

#### Question 8

The polynomials ax^{3} + 3x^{2} – 3 and 2x^{3} – 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.

#### Answer 8

Let f(x) = ax^{3} + 3x^{2} – 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) = 2x^{3} – 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

#### Question 9

Find the value of ‘a’, if (x – a) is a factor of x^{3} – ax^{2} + x + 2.

#### Solution 9

Let f(x) = x^{3} – ax^{2} + x + 2

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

Remainder = f(a) = 0

a^{3} – a^{3} + a + 2 = 0

a + 2 = 0

a = -2

#### Question 10

Find the number that must be subtracted from the polynomial 3y^{3} + y^{2} – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.

#### Solution 10

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

Let f(y) = 3y^{3} + y^{2} – 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 Maths Solution Exercise – 8(C) Remainder And Factor Theorems**

#### Question 1

Show that (x – 1) is a factor of x^{3} – 7x^{2} + 14x – 8. Hence, completely factorise the given expression.

#### Answer 1

Let f(x) = x^{3} – 7x^{2} + 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).

#### Question 2

Using Remainder Theorem, factorise:

x^{3} + 10x^{2} – 37x + 26 completely.

#### Answer 2

#### Question 3

When x^{3} + 3x^{2} – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.

#### Answer 3

Let f(x) = x^{3} + 3x^{2} – 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

#### Question 4

What should be subtracted from 3x^{3} – 8x^{2} + 4x – 3, so that the resulting expression has x + 2 as a factor?

#### Answer 4

Let the required number be k.

Let f(x) = 3x^{3} – 8x^{2} + 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.

#### Question 5

If (x + 1) and (x – 2) are factors of x^{3} + (a + 1)x^{2} – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.

#### Answer 5

Let f(x) = x^{3} + (a + 1)x^{2} – (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) = x^{3} + 2x^{2} – 5x – 6

Now, (x + 1) and (x – 2) are factors of f(x). Hence, (x + 1) (x – 2) = x^{2} – x – 2 is a factor of f(x).

f(x) = x^{3} + 2x^{2} – 5x – 6 = (x + 1) (x – 2) (x + 3)

#### Question 6

If x – 2 is a factor of x^{2} + ax + b and a + b = 1, find the values of a and b.

#### Answer 6

Let f(x) = x^{2} + 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

#### Question 7

Factorise x^{3} + 6x^{2} + 11x + 6 completely using factor theorem.

#### Answer 7

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

#### Question 8

Find the value of ‘m’, if mx^{3} + 2x^{2} – 3 and x^{2} – mx + 4 leave the same remainder when each is divided by x – 2.

#### Answer 8

Let f(x) = mx^{3} + 2x^{2} – 3

g(x) = x^{2} – 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 =3/8

#### Question 9

The polynomial px^{3} + 4x^{2} – 3x + q is completely divisible by x^{2} – 1; find the values of p and q. Also, for these values of p and q factorize the given polynomial completely.

#### Answer 9 **Remainder And Factor Theorems**

Let f(x) = px^{3} + 4x^{2} – 3x + q

It is given that f(x) is completely divisible by (x^{2} – 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) = 3x^{3} + 4x^{2} – 3x – 4

Given that f(x) is completely divisible by (x^{2} – 1).

#### Question 10

Find the number which should be added to x^{2} + x + 3 so that the resulting polynomial is completely divisible by (x + 3).

#### Answer 10

Let the required number be k.

Let f(x) = x^{2} + 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.

#### Question 11 **Remainder And Factor Theorems**

When the polynomial x^{3} + 2x^{2} – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x^{3} + ax^{2} – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.

#### Answer 11

It is given that when the polynomial x^{3} + 2x^{2} – 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 x^{3} + ax^{2} – 12x + 16 is divided by (x + 2), the remainder is B.

x^{3} + ax^{2} – 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

#### Question 12

(3x + 5) is a factor of the polynomial (a – 1)x^{3} + (a + 1)x^{2} – (2a + 1)x – 15. Find the value of ‘a’, factorise the given polynomial completely.

#### Answer 12

Let f(x) = (a – 1)x^{3} + (a + 1)x^{2} – (2a + 1)x – 15

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

f(x) = (a – 1)x^{3} + (a + 1)x^{2} – (2a + 1)x – 15

= 3x^{3} + 5x^{2} – 9x – 15

#### Question 13 **Remainder And Factor Theorems**

When divided by x – 3 the polynomials x^{3} – px^{2} + x + 6 and 2x^{3} – x^{2} – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’.

#### Answer 3

If (x – 3) divides f(x) = x^{3} – px^{2} + x + 6, then,

Remainder = f(3) = 3^{3} – p(3)^{2} + 3 + 6 = 36 – 9p

If (x – 3) divides g(x) = 2x^{3} – x^{2} – (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

#### Question 14 **Remainder And Factor Theorems**

Use the Remainder Theorem to factorise the following expression:

2x^{3} + x^{2} – 13x + 6

#### Answer 14

f(x) = 2x^{3} + x^{2} – 13x + 6

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

Putting x = 2, we have:

f(2) = 2(2)^{3} + 2^{2} – 13 (2) + 6 = 16 + 4 – 26 + 6 = 0

Hence (x – 2) is a factor of f(x).

#### Question 15

Using remainder theorem, find the value of k if on dividing 2x^{3} + 3x^{2} – kx + 5 by x – 2, leaves a remainder 7.

#### Answer 15

Let f(x) = 2x^{3} + 3x^{2} – kx + 5

Using Remainder Theorem, we have

f(2) = 7

∴ 2(2)^{3} + 3(2)^{2} – k(2) + 5 = 7

and ∴ 16 + 12 – 2k + 5 = 7

so ∴ 33 – 2k = 7

Therefor ∴ 2k = 26

Hence ∴ k = 13

#### Question 16

What must be subtracted from 16x^{3} – 8x^{2} + 4x + 7 so that the resulting expression has 2x + 1 as a factor?

#### Answer 16

Here, f(x) = 16x^{3} – 8x^{2} + 4x + 7

Let the number subtracted be k from the given polynomial f(x).

Given that 2x + 1 is a factor of f(x).

Therefore 1 must be subtracted from 16x^{3} – 8x^{2} + 4x + 7 so that the resulting expression has 2x + 1 as a factor.

**———-End of Remainder And Factor Theorems—– **

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**Chapter Wise Concise Maths Solution for ICSE Class 10th**

**GST ( Goods and Services Tax)****Banking****Shares and Dividends****Linear****In equations****in One Variable****Quadratic Equations****Solving Simple Problems (Based on Quadratic Equations)****Ratio and Proportion ( Including Properties and Uses )****Reminder and Factor Theorems (Currently Open )****Matrices****Arithmetic Progression****Geometric Progression****Reflection****Section And Mid Point Formula****Equation of a Line****Similarity (With Applications to Maps and Models )****Loci (Locus and its Constructions)****Circles****Tangents and Intersecting Chords****Constructions ( Circles)****Cylinder, Cone and Spheres (Surface Area and Volume )****Trigonometrical Identities****Heights and Distances****Graphical Representation (Histograms and Ogives)****Measures of Central Tendency (Mean , Median, Quartiles and Mode)****Probability**