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ICSE Class 9 Maths Selina Solutions Chapter 5 Factorisation

ICSE Class 9 Maths Selina Solutions Chapter 5 are provided here which have been prepared by our PW experts. Get the details of the ICSE Class 9 Maths Selina Solutions Chapter 5 - Factorisation Overview in the below article.
authorImagePriyanka Dahima16 Jul, 2024
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ICSE Class 9 Maths Selina Solutions Chapter 5

ICSE Class 9 Maths Selina Solutions Chapter 5: Here we provide you with detailed ICSE Class 9 Maths Selina Solution Chapter 5 Factorization. Our PW Subject experts have prepared these questions  according to the latest syllabus for ICSE Class 9 Maths exam.

Selina's solutions are considered very useful for ICSE Class 9 Maths exam preparation. For better and more strategic preparation in the right direction, our blog serves as a guide for ICSE Class 9 Maths 2024 aspirants.

ICSE Class 9 Maths Selina Solutions Chapter 5

In mathematics, Factorization is writing a number or other mathematical object as the product of several factors, usually the product of smaller or simpler similar objects. For example, 3 × 5 is an integer factorization of 15 and is a polynomial factorization of x²- 4.
Selina's solutions are considered very useful for ICSE class 9 Maths Selina Solutions Chapter 5 preparation. Here we bring you with detailed answers and solutions for Selina Solutions for Class 9 Maths Chapter 5 - Factorization. Subject experts have prepared these questions as per the syllabus prescribed by CISCE for ICSE.

ICSE Class 9 Maths Selina Solutions Chapter 5 PDF

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ICSE Class 9 Maths Selina Solutions Chapter 5 PDF

ICSE Class 9 Maths Selina Solutions Chapter 5 Factorization

Here we have provided ICSE Class 9 Maths Selina Solutions Chapter 5 Factorization for the ease of students so that they can prepare better for their ICSE Class 9 Maths Exam.

ICSE Class 9 Maths Selina Solutions Chapter 5 Exercise 5(A)

Factorise by taking out the common factors:

1. 2(2x – 5y) (3x + 4y) – 6(2x – 5y) (x – y)

Solution:

Identifying and taking (2x – 5y) common from both the terms, we have = (2x – 5y) [2(3x + 4y) – 6(x – y)] = (2x – 5y) (6x + 8y – 6x + 6y) = (2x – 5y) (8y + 6y) = (2x – 5y) (14y) = (2x – 5y)14y

2. xy(3x 2 – 2y 2 ) – yz(2y 2 – 3x 2 ) + zx(15x 2 – 10y 2 )

Solution:

We have, xy(3x 2 – 2y 2 ) – yz(2y 2 – 3x 2 ) + zx(15x 2 – 10y 2 ) Changing signs to arrive at a common term So, = xy(3x 2 – 2y 2 ) + yz(3x 2 – 2y 2 ) + zx(15x 2 – 10y 2 ) = xy(3x 2 – 2y 2 ) + yz(3x 2 – 2y 2 ) + 5zx(3x 2 – 2y 2 ) = (3x 2 – 2y 2 ) (xy + yz + 5zx)

3. ab(a 2 + b 2 – c 2 ) – bc(c 2 – a 2 – b 2 ) + ca(a 2 + b 2 – c 2 )

Solution:

We have, ab(a 2 + b 2 – c 2 ) – bc(c 2 – a 2 – b 2 ) + ca(a 2 + b 2 – c 2 ) Changing signs to arrive at a common term So, = ab(a 2 + b 2 – c 2 ) + bc(a 2 + b 2 – c 2 ) + ca(a 2 + b 2 – c 2 ) = (a 2 + b 2 – c 2 ) (ab + bc + ca)

4. 2x(a – b) + 3y(5a – 5b) + 4z(2b – 2a)

Solution:

We have, 2x(a – b) + 3y(5a – 5b) + 4z(2b – 2a) Taking common factors, we get = 2x(a – b) + 15y(a – b) – 8z(a – b) = (a – b) (2x + 15y – 8z)

Factorize by the grouping method:

5. a 3 + a – 3a 2 – 3

Solution:

We have, a 3 + a – 3a 2 – 3 Grouping to arrive at a common term = a (a 2 + 1) – 3(a 2 + 1) Taking common, we get = (a 2 + 1) (a -3)

6. 16(a + b) 2 – 4a – 4b

Solution:

We have, 16 (a + b) 2 – 4a – 4b Grouping to arrive at a common term = 16(a + b) 2 – 4 (a + b) Taking common, we get = 4(a + b) [4 (a + b) – 1] = 4(a + b) (4a + 4b – 1)

7. Factorize by the grouping method:

a 4 – 2a 3 – 4a + 8

Solution:

We have, a 4 – 2a 3 – 4a + 8 Grouping to arrive at a common term = a 3 (a – 2) – 4(a – 2) Taking common, we get = (a 3 – 4) (a – 2)

8. ab – 2b + a 2 – 2a

Solution:

We have, ab – 2b + a 2 – 2a Grouping to arrive at a common term = b(a – 2) + a(a – 2) Taking common, we get = (b + a) (a – 2)

9. ab(x 2 + 1) + x(a 2 + b 2 )

Solution:

We have, ab(x 2 + 1) + x(a 2 + b 2 ) On expanding, = abx 2 + ab + a 2 x + b 2 x Now, grouping to arrive at a common term = abx 2 + a 2 x + b 2 x + ab = ax(bx + a) + b(bx + a) Taking common, we get = (ax + b) (bx + a)

10. a 2 + b – ab – a

Solution:

We have, a 2 + b – ab – a Grouping to arrive at a common term = a 2 – a + b – ab = a(a – 1) – b(-1 + a) = a(a – 1) – b(a – 1) Taking common, we get = (a – b) (a – 1)

11. (ax + by) 2 + (bx – ay) 2

Solution:

We have, (ax + by) 2 + (bx – ay) 2 On expanding, = a 2 x 2 + b 2 y 2 + 2abxy + b 2 x 2 + a 2 y 2 – 2abxy = a 2 x 2 + b 2 y 2 + b 2 x 2 + a 2 y 2 Rearranging terms, we get = a 2 x 2 + b 2 x 2 + a 2 y 2 + b 2 y 2 Taking common, we get = x 2 (a 2 + b 2 ) + y 2 (a 2 + b 2 ) = (x 2 + y 2 ) (a 2 + b 2 )

12. a 2 x 2 + (ax 2 + 1)x + a

Solution:

We have, a 2 x 2 + (ax 2 + 1)x + a Regrouping the terms, we have = a 2 x 2 + a + (ax 2 + 1)x = a(ax 2 + 1) + x(ax 2 + 1) Taking common, we get = (ax 2 + 1) (a + x)

13. (2a – b) 2 – 10a + 5b

Solution:

We have, (2a – b) 2 – 10a + 5b Taking common, = (2a – b) 2 – 5(2a – b) Now, = (2a – b) [(2a – b) – 5] = (2a – b) (2a – b – 5)

14. a(a – 4) – a + 4

Solution:

We have, a(a – 4) – a + 4 By grouping, we get = a(a – 4) -1(a – 4) Now, taking the common term = (a – 4) (a – 1)

15. y 2 – (a + b) y + ab

Solution:

We have, y 2 – (a + b) y + ab On expanding, = y 2 – ay – by + ab = (y 2 – ay) – by + ab Taking ‘y’ and ‘b’ common from the group, we get = y(y – a) – b(y – a) = (y – a) (y – b)

16. a 2 + 1/a 2 – 2 – 3a + 3/a Solution:

We have, a 2 + 1/a 2 – 2 – 3a + 3/a On grouping terms, we get = (a 2 – 2 + 1/a 2 ) – 3a + 3/a = [a 2 – (2 x a x 1/a) + 1/a 2 ] – 3(a – 1/a) = (a – 1/a) 2 – 3(a – 1/a) {Since, (x – y) 2 = x 2 – 2xy + y 2 } Taking (a – 1/a) as common, we get = (a – 1/a) [(a – 1/a) – 3] = (a – 1/a) (a – 1/a – 3)

17. x 2 + y 2 + x + y + 2xy

Solution:

We have, x 2 + y 2 + x + y + 2xy On rearranging terms, we get = (x 2 + y 2 + 2xy) + (x + y) {Since, (x + y) 2 = x 2 + 2xy + y 2 } Now, = (x + y) 2 + (x + y) = (x + y)(x + y + 1)

18. a 2 + 4b 2 – 3a + 6b – 4ab

Solution:

We have, a 2 + 4b 2 – 3a + 6b – 4ab On rearranging terms, we get = a 2 + 4b 2 – 4ab – 3a + 6b Now, = a 2 + (2b) 2 – 2 × a × (2b) – 3(a – 2b) {Since, (a – b) 2 = a 2 – 2ab + b 2 } = (a – 2b) 2 – 3(a – 2b) = (a – 2b) [(a – 2b)- 3] = (a – 2b) (a – 2b – 3)

19. m (x – 3y) 2 + n (3y – x) + 5x – 15y

Solution:

We have, m (x – 3y) 2 + n (3y – x) + 5x – 15y Now, Taking (x – 3y) common from all the three terms, we get = m(x – 3y) 2 – n(x – 3y) + 5(x – 3y) = (x – 3y) [m(x – 3y) – n + 5] = (x – 3y) (mx – 3my – n + 5)

20. x(6x – 5y) – 4(6x – 5y) 2

Solution:

We have, x(6x – 5y) – 4(6x – 5y) 2 Now, Taking (6x – 5y) common from the three terms, we get = (6x – 5y) [x – 4(6x – 5y)] = (6x – 5y) (x – 24x + 20y) = (6x – 5y) (-23x + 20y) = (6x – 5y) (20y – 23x)

Exercise 5(B)

Factorize:

1. a 2 + 10a + 24

Solution:

We have, a 2 + 10a + 24 By splitting the middle term, we get = a 2 + 6a + 4a + 24 = a(a + 6) + 4(a + 6) = (a + 4) (a + 6)

2. a 2 – 3a – 40

Solution:

We have, a 2 – 3a – 40 By splitting the middle term, we get = a 2 – 8a + 5a – 40 = a(a – 8) + 5(a – 8) = (a + 5) (a – 8)

3. 1 – 2a – 3a 2

Solution:

We have, 1 – 2a – 3a 2 By splitting the middle term, we get = 1 – 3a + a – 3a 2 = 1(1 – 3a) + a(1 – 3a) = (1 + a) (1 – 3a)

4. x 2 – 3ax – 88a 2

Solution:

We have, x 2 – 3ax – 88a 2 By splitting the middle term, we get = x 2 – 11ax + 8ax – 88a 2 = x(x – 11a) + 8a(x – 11a) = (x + 8a) (x – 11a)

5. 6a 2 – a – 15

Solution:

We have, 6a 2 – a – 15 By splitting the middle term, we get = 6a 2 + 9a – 10a – 15 = 3a(2a + 3) – 5(2a + 3) = (3a – 5) (2a + 3)

6. 24a 3 + 37a 2 – 5a

Solution:

We have, 24a 3 + 37a 2 – 5a Taking ‘a’ common from all = a (24a 2 + 37a – 5) = a (24a 2 + 40a – 3a – 5) {By splitting the middle term} = a [8a(3a + 5) – 1(3a + 5)] = a [(8a – 1) (3a + 5)] = a (8a – 1) (3a + 5)

7. a(3a – 2) – 1

Solution:

We have, a(3a – 2) – 1 On expanding, = 3a 2 – 2a – 1 By splitting the middle term, we get = 3a 2 – 3a + a – 1 = 3a(a – 1) + 1(a – 1) = (3a + 1) (a – 1)

8. a 2 b 2 + 8ab – 9

Solution:

We have, a 2 b 2 + 8ab – 9 By splitting the middle term, we get = a 2 b 2 + 9ab – ab – 9 = ab(ab + 9) – 1(ab + 9) = (ab – 1) (ab + 9)

9. 3 – a (4 + 7a)

Solution:

We have, 3 – a (4 + 7a) On expanding, = 3 – 4a – 7a 2 By splitting the middle term, we get = 3 + 3a – 7a – 7a 2 = 3(1 + a) – 7a(1 + a) = (1 + a) (3 – 7a)

10. (2a + b) 2 – 6a – 3b – 4

Solution:

We have, (2a + b) 2 – 6a – 3b – 4 = (2a + b) 2 – 3(2a + b) – 4 Let’s assume that (2a + b) = x So, the expression becomes = x 2 – 3x – 4 By splitting the middle term, we get = x 2 – 4x + x – 4 = x(x – 4) + 1(x – 4) = (x – 4) (x + 1) Resubstituting the value of x, we get = (2a + b – 4) (2a + b + 1)

11. 1 – 2 (a+ b) – 3 (a + b) 2

Solution:

We have, 1 – 2 (a+ b) – 3 (a + b) 2 Let’s assume (a + b) = x Then, the expression becomes = 1 – 2x – 3x 2 By splitting the middle term, we get = 1 – 3x + x – 3x 2 = 1(1 – 3x) + x(1 – 3x) = (1 – 3x) (1 + x) Resubstituting the value of x, we get = [1 – 3(a + b)] [1 + (a + b)] = (1 – 3a – 3b) (1 + a + b)

12. 3a 2 – 1 – 2a

Solution:

We have, 3a 2 – 1 – 2a Rearranging, = 3a 2 – 2a – 1 By splitting the middle term, we get = 3a 2 – 3a + a – 1 = 3a(a – 1) + 1(a – 1) = (3a + 1) (a – 1)

13. x 2 + 3x + 2 + ax + 2a

Solution:

We have, x 2 + 3x + 2 + ax + 2a By splitting the middle term, we get = (x 2 + 2x + x + 2) + ax + 2a = x(x + 2) + 1(x + 2) + a(x + 2) = (x + 2) (x + a + 1)

14. (3x – 2y) 2 + 3 (3x – 2y) – 10

Solution:

We know, (3x – 2y) 2 + 3 (3x – 2y) – 10 Let’s assume that (3x – 2y) = a So, the expression becomes = a 2 + 3a – 10 By splitting the middle term, we get = a 2 + 5a – 2a – 10 = a(a + 5) – 2(a + 5) = (a – 2) (a + 5) = (3x – 2y + 5) (3x – 2y – 2)

15. 5 – (3a 2 – 2a) (6 – 3a 2 + 2a)

Solution:

Given, 5 – (3a 2 – 2a) (6 – 3a 2 + 2a) = 5 – (3a 2 – 2a) [6 – (3a 2 – 2a)] Let’s substitute (3a 2 – 2a) = x And, the expression becomes, = 5 – x(6 – x) = 5 – 6x – x 2 = 5 – 5x – x – x 2 = 5(1 – x) – x(1 – x) = (1 – x) (5 – x) = (x – 1) (x – 5) = (3a 2 – 2a – 1) (3a 2 – 2a – 5) Now, = (3a 2 – 3a + a – 1) (3a 2 + 3a – 5a – 5) {By splitting the middle term} = [3a(a – 1) + 1(a – 1)] [3a(a + 1) – 5(a + 1)] = [(3a + 1) (a – 1)] [(3a – 5) (a + 1)] = (3a + 1) (3a – 5) (a + 1)(a – 1)

16. 1/35 + 12a/35 + a 2

Solution:

We have, 1/35 + 12a/35 + a 2 Taking common, = 1/35 (1 + 12a + 35a 2 ) = 1/35 (35a 2 + 12a + 1) = 1/35 (35a 2 + 7a + 5a + 1) {By splitting the middle term} = 1/35 [7a(5a + 1) + 1(5a + 1)] = 1/35 [(7a + 1) (5a + 1)] = [(7a + 1) (5a + 1)]/ 35

17. (x 2 – 3x) (x 2 – 3x – 1) – 20.

Solution:

We have, (x 2 – 3x) (x 2 – 3x – 1) – 20 = (x 2 – 3x)[(x 2 – 3x) – 1] – 20 Let’s = a[a – 1] – 20 ….(Taking x 2 – 3x = a) = a 2 – a – 20 = a 2 – 5a + 4a – 20 = a(a – 5) + 4(a – 5) = (a – 5)(a + 4) = (x 2 – 3x – 5)(x 2 – 3x + 4)

18. Find each trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.

(i) x 2 – 3x – 54

(ii) 2x 2 – 7x – 15

(iii) 2x 2 + 2x – 75

(iv) 3x 2 + 4x – 10

(v) x(2x – 1) – 1

Solution:

(i) Given, x 2 – 3x – 54 On comparing with the general form ax 2 + bx + c, we get a = 1, b = -3 and c = -54 So, b 2 – 4ac = (-3) 2 – 4(1)(-54) = 9 + 216 = 225 225 is a perfect square Thus, x 2 – 3x – 54 is factorisable Now, x 2 – 3x – 54 = x 2 – 9x + 6x – 54 = x(x – 9) + 6(x – 9) = (x + 6) (x – 9) (ii) Given, 2x 2 – 7x – 15 On comparing with the general form ax 2 + bx + c, we get a = 2, b = -7 and c = -15 So, b 2 – 4ac = (-7) 2 – 4(2)(-15) = 49 + 120 = 169 169 is a perfect square Thus, 2x 2 – 7x – 15 is factorisable Now, 2x 2 – 7x – 15 = 2x 2 – 10x + 3x – 15 = 2x(x – 5) + 3(x – 5) = (2x + 3) (x – 5) (iii) Given, 2x 2 + 2x – 75 On comparing with the general form ax 2 + bx + c, we get a = 2, b = 2 and c = -75 So, b 2 – 4ac = (2) 2 – 4(2)(-75) = 4 + 600 = 604 604 is not a perfect square Thus, 2x 2 + 2x – 75 is not factorizable (iv) Given, 3x 2 + 4x – 10 On comparing with the general form ax 2 + bx + c, we get a = 3, b = 4 and c = -10 So, b 2 – 4ac = (4) 2 – 4(3)(-10) = 16 + 120 = 136 136 is a not perfect square Thus, 3x 2 + 4x – 10 is not factorizable (v) Given, x(2x – 1) – 1 = 2x 2 – x – 1 On comparing with the general form ax 2 + bx + c, we get a = 2, b = -1 and c = -1 So, b 2 – 4ac = (-1) 2 – 4(2)(-1) = 1 + 8 = 9 9 is a perfect square Thus, x(2x – 1) – 1 is factorisable Now, x(2x – 1) – 1 = 2x 2 – x – 1 = 2x 2 – 2x + x – 1 = 2x(x – 1) + 1(x – 1) = (2x + 1) (x – 1)

19. Factorise:

(i) 4√3x 2 + 5x – 2√3

(ii) 7√2x 2 – 10x – 4√2

Solution:

(i) We have, 4√3x 2 + 5x – 2√3 By splitting the middle term, we get = 4√3x 2 + 8x – 3x – 2√3 = 4x(√3x + 2) – √3(√3x + 2) = (4x – √3) (√3x + 2) (ii) We have, 7√2x 2 – 10x – 4√2 By splitting the middle term, we get = 7√2x 2 – 14x + 4x – 4√2 = 7√2x(x – √2) + 4(x – √2) = (7√2x + 4) (x – √2)

20. Give possible expressions for the length and the breadth of the rectangle whose area is 12x 2 – 35x + 25.

Solution:

We have, 12x 2 – 35x + 25 By splitting the middle term, we get = 12x 2 – 20x – 15x + 25 = 4x(3x – 5) – 5(3x – 5) = (3x – 5) (4x – 5) Hence, Length = (3x – 5) and breadth = (4x – 5) or, Length = (4x – 5) and breadth = (3x – 5)

Exercise 5(C)

Factorize:

1. 25a 2 – 9b 2

Solution:

We have, 25a 2 – 9b 2 = (5a) 2 – (3b) 2 = (5a + 3b) (5a – 3b) [As x 2 – y 2 = (x + y)(x – y)

2. a 2 – (2a + 3b) 2

Solution:

We have, a 2 – (2a + 3b) 2 = [a – (2a + 3b)] [a + (2a + 3b)] [As x 2 – y 2 = (x + y)(x – y)] = (a – 2a – 3b) (a + 2a + 3b) = (-a – 3b) (3a + 3b) = -3(a + 3b) (a + b)

3. a 2 – 81(b-c) 2

Solution:

We have, a 2 – 81(b-c) 2 = a 2 – [9(b – c)] 2 = [a – 9(b – c)] [a + 9(b – c)]] [As x 2 – y 2 = (x + y)(x – y)] = (a – 9b + 9c) (a + 9b – 9c)

4. 25(2a – b) 2 – 81b 2

Solution:

We have, 25(2a – b) 2 – 81b 2 = [5(2a – b)] 2 – (9b) 2 = [5(2a – b) – 9b] [5(2a – b) + 9b] [As x 2 – y 2 = (x + y)(x – y)] = (10a – 5b – 9b) (10a – 5b + 9b) = (10a – 14b) (10a + 4b) = 2(5a – 7b). 2(5a + 2b) = 2(5a – 7b) (5a + 2b)

5. 50a 3 – 2a

Solution:

We have, 50a 3 – 2a = 2a(25a 2 – 1) = 2a[(5a) 2 – 1 2 ] = 2a(5a – 1)(5a + 1) [As x 2 – y 2 = (x + y)(x – y)

6. 4a 2 b – 9b 3

Solution:

We have, 4a 2 b – 9b 3 = b(4a 2 – 9b 2 ) = b[(2a) 2 – (3b) 2 ] = b[(2a + 3b) (2a – 3b)] [As x 2 – y 2 = (x + y)(x – y)] = b(2a + 3b)(2a – 3b)

7. 3a 5 – 108a 3

Solution:

We have, 3a 5 – 108a 3 = 3a 3 (a 2 – 36) = 3a 3 (a 2 – 6 2 ) = 3a 3 (a – 6) (a + 6) [As x 2 – y 2 = (x + y)(x – y)]

8. 9(a – 2) 2 – 16(a + 2) 2

Solution:

We have, 9(a – 2) 2 – 16(a + 2) 2 = [3(a – 2)] 2 – [4(a + 2)] 2 = [3(a – 2) – 4(a + 2)] [3(a – 2) + 4(a + 2)] [As x 2 – y 2 = (x + y)(x – y)] = [3a – 6 – 4a – 8] [3a – 6 + 4a + 8] = [-a – 14] [7a + 2] = -(a + 14) (7a + 2)

9. a 4 – 1

Solution:

We have, a 4 – 1 = (a 2 ) 2 – 1 2 = (a 2 – 1) (a 2 + 1) [As x 2 – y 2 = (x + y)(x – y)] = [(a – 1)(a + 1)] (a 2 + 1) = (a – 1)(a + 1)(a 2 + 1)

10. a 3 + 2a 2 – a – 2

Solution:

We have, a 3 + 2a 2 – a – 2 = a 2 (a + 2) – 1(a + 2) = (a 2 – 1) (a + 2) = (a – 1) (a + 1) (a + 2) [As x 2 – y 2 = (x + y)(x – y)]

11. (a + b) 3 – a – b

Solution:

We have, (a + b) 3 – a – b = (a + b) 3 – (a + b) = (a + b) [(a + b) 2 – 1] = (a + b) [(a + b – 1) (a + b + 1)] [As x 2 – y 2 = (x + y)(x – y)] = (a + b) (a + b -1) (a + b + 1)

12. a(a – 1) – b(b – 1)

Solution:

We have, a(a – 1) – b(b – 1) = a 2 – a – b 2 + b = (a 2 – b 2 ) – (a – b) = (a + b) (a – b) – (a – b) [As x 2 – y 2 = (x + y)(x – y)] = (a – b) [(a + b) – 1] = (a – b) (a + b – 1)

13. 4a 2 – (4b 2 + 4bc + c 2 )

Solution:

We know, 4a 2 – (4b 2 + 4bc + c 2 ) = (2a) 2 – [(2b) 2 + 2(2b)(c) + c 2 ] = (2a) 2 – (2b + c) 2 = (2a – 2b – c) (2a + 2b + c) [As x 2 – y 2 = (x + y)(x – y)]

14. 4a 2 – 49b 2 + 2a – 7b

Solution:

We know, 4a 2 – 49b 2 + 2a – 7b = (2a) 2 – (7b) 2 + (2a – 7b) = [(2a – 7b) (2a + 7b)] + (2a – 7b) [As x 2 – y 2 = (x + y)(x – y)] = (2a – 7b) [(2a + 7b) + 1] = (2a – 7b) (2a + 7b + 1)

15. 9a 2 + 3a – 8b – 64b 2

Solution:

We have, 9a 2 + 3a – 8b – 64b 2 = 9a 2 – 64b 2 + 3a – 8b = (3a) 2 – (8b) 2 + (3a – 8b) = [(3a – 8b) (3a + 8b)] + (3a – 8b) [As x 2 – y 2 = (x + y)(x – y)] = (3a – 8b) [(3a + 8b) + 1] = (3a – 8b) (3a + 8b + 1)

16. 4a 2 – 12a + 9 – 49b 2

Solution:

We have, 4a 2 – 12a + 9 – 49b 2 = [(2a) 2 – 2(2a)(3) + 3 2 ] – (7b) 2 = (2a – 3) 2 – (7b) 2 = (2a – 7b – 3) (2a + 7b – 3) [As x 2 – y 2 = (x + y)(x – y)]

17. 4xy – x 2 – 4y 2 + z 2

Solution:

We have, 4xy – x 2 – 4y 2 + z 2 On rearranging, = z 2 – x 2 – 4y 2 + 4xy = z 2 – (x 2 + 4y 2 – 4xy) = z 2 – (x – 2y) 2 = (z – x + 2y) (z + x – 2y) [As x 2 – y 2 = (x + y)(x – y)]

18. a 2 + b 2 – c 2 – d 2 + 2ab – 2cd

Solution:

We have, a 2 + b 2 – c 2 – d 2 + 2ab – 2cd On rearranging, = a 2 + 2ab + b 2 – c 2 – d 2 – 2cd = (a 2 + 2ab + b 2 ) – (c 2 + d 2 + 2cd) = (a + b) 2 – (c + d) 2 = (a + b + c + d) (a + b – c – d) [As x 2 – y 2 = (x + y)(x – y)]

19. 4x 2 – 12ax – y 2 – z 2 – 2yz + 9a 2

Solution:

We have, 4x 2 – 12ax – y 2 – z 2 – 2yz + 9a 2 On rearranging, = 4x 2 – 12ax + 9a 2 – y 2 – z 2 – 2yz = (4x 2 – 12ax + 9a 2 ) – (y 2 + z 2 + 2yz) = (2x – 3a) 2 – (y + z) 2 = (2x – 3a + y + z) (2x – 3a – y – z) [As x 2 – y 2 = (x + y)(x – y)]

20. (a 2 – 1) (b 2 – 1) + 4ab

Solution:

We have, (a 2 – 1) (b 2 – 1) + 4ab By cross multiplying and expanding, we get = (1 – a 2 – b 2 + a 2 b 2 ) + 4ab On manipulating, = (a 2 b 2 + 1 + 2ab) – (a 2 + b 2 – 2ab) Now, = (ab + 1) 2 – (a – b) 2 = [(ab + 1) – (a – b)] [(ab + 1) + (a – b)] = (ab + 1 – a + b) (ab + 1 + a – b) [As x 2 – y 2 = (x + y)(x – y)]

21. x 4 + x 2 + 1

Solution:

We have, x 4 + x 2 + 1 = x 4 + 2x 2 + 1 – x 2 = (x 2 + 1) 2 – x 2 [As a 2 – b 2 = (a + b)(a – b)] = (x 2 + 1 – x) (x 2 + 1 + x)

22. (a 2 + b 2 – 4c 2 ) 2 – 4a 2 b 2

Solution:

We have, (a 2 + b 2 – 4c 2 ) 2 – 4a 2 b 2 = (a 2 + b 2 – 4c 2 ) 2 – (2ab) 2 = [(a 2 + b 2 – 4c 2 ) + (2ab)] [(a 2 + b 2 – 4c 2 ) – (2ab)] [As x 2 – y 2 = (x + y)(x – y)] = [(a 2 + b 2 + 2ab) – 4c 2 ] [(a 2 + b 2 – 2ab) – 4c 2 ] = [(a + b) 2 – (2c) 2 ] [(a – b) 2 – (2c) 2 ] = [(a + b – 2c) (a + b + 2c)] [(a – b – 2c) (a – b + 2c)] [As x 2 – y 2 = (x + y)(x – y)] = (a + b – 2c) (a + b + 2c) (a – b – 2c) (a – b + 2c)

23. (x 2 + 4y 2 – 9z 2 ) 2 – 16x 2 y 2

Solution:

We have, (x 2 + 4y 2 – 9z 2 ) 2 – 16x 2 y 2 = (x 2 + 4y 2 – 9z 2 ) 2 – (4xy) 2 = (x 2 + 4y 2 – 9z 2 – 4xy) (x 2 + 4y 2 – 9z 2 + 4xy) [As x 2 – y 2 = (x + y)(x – y)] = [(x 2 – 4xy + 4y 2 ) – 9z 2 ] [(x 2 + 4xy + 4y 2 ) – 9z 2 ] = [(x – 2y) 2 – (3z) 2 ] [(x + 2y) 2 – (3z) 2 ] = [(x – 2y + 3z) (x – 2y – 3z)] [(x +2y + 3z) (x + 2y – 3z)] [As x 2 – y 2 = (x + y)(x – y)] = (x – 2y + 3z) (x – 2y – 3z)] [(x +2y + 3z) (x + 2y – 3z)

24. (a + b) 2 – a 2 + b 2

Solution:

We have, (a + b) 2 – a 2 + b 2 On expanding, = (a 2 + 2ab + b 2 ) – a 2 + b 2 = 2b 2 + 2ab = 2b (b + a)

25. a 2 – b 2 – (a + b) 2

Solution:

We have, a 2 – b 2 – (a + b) On expanding, = a 2 – b 2 – (a 2 + b 2 + 2ab) = a 2 – b 2 – a 2 – b 2 – 2ab = -2b 2 – 2ab = -2b(b + a)

26. 9a 2 – (a 2 – 4) 2

Solution:

We have, 9a 2 – (a 2 – 4) 2 = (3a) 2 – (a 2 – 4) 2 = [3a – (a 2 – 4)] [3a + (a 2 – 4)] [As x 2 – y 2 = (x + y)(x – y)] = [3a – (a 2 – 2 2 )] [3a + (a 2 – 2 2 )] = (3a – a 2 + 4) (3a + a 2 – 4) = (-a 2 + 3a + 4) (a 2 + 3a – 4) = (-a 2 + 4a – a + 4) (a 2 + 4a – a – 4) [By splitting the middle term] = [a(-a + 4) + 1(-a + 4)] [a(a + 4) – 1(a + 4)] = [(-a + 4) (a + 1)] [(a – 1) (a + 4)] = (4 – a) (a + 1) (a – 1) (a + 4)

27. x 2 + 1/x 2 – 11

Solution:

We have, x 2 + 1/x 2 – 11 = x 2 + 1/x 2 – 2 – 9 = (x 2 + 1/x 2 – 2 × x × 1/x) – 9 = (x – 1/x) 2 – 3 2 = (x – 1/x + 3) (x – 1/x – 3)  [As a 2 – b 2 = (a + b)(a – b)]

28. 4x 2 + 1/4x 2 + 1

Solution:

We have, 4x 2 + 1/4x 2 + 1 = 4x 2 + 1/4x 2 + 2 – 1 = [(2x) 2 + (1/2x) 2 + 2 × 2x × 1/2x)] – 1 2 = (2x + 1/2x) 2 – 1 2 = (2x + 1/2x + 1) (2x – 1/2x – 1)  [As x 2 – y 2 = (x + y)(x – y)]

29. 4x 4 – x 2 – 12x – 36

Solution:

We know, 4x 4 – x 2 – 12x – 36 = 4x 4 – (x 2 + 12x + 36) = (2x 2 ) 2 – [x 2 + 2(x)(6) + 6 2 ] = (2x 2 ) 2 – (x + 6) 2 = (2x 2 + x + 6) (2x 2 – x – 6) [As x 2 – y 2 = (x + y)(x – y)] = (2x 2 + x + 6) (2x 2 – 4x + 3x – 6) [By splitting the middle term] = (2x 2 + x + 6) [2x(x – 2) + 3(x – 2)] = (2x 2 + x + 6) [(2x + 3) (x – 2)] = (2x 2 + x + 6) (2x + 3) (x – 2)

30. a 2 (b + c) – (b + c) 3

Solution:

We have, a 2 (b + c) – (b + c) 3 = (b + c) [a 2 – (b + c) 2 ] = (b + c) [(a – b – c) (a + b + c)] [As x 2 – y 2 = (x + y)(x – y)] = (b + c) (a – b – c) (a + b + c)

Exercise 5(D)

Factorize:

1. a 3 – 27

Solution:

We have, a 3 – 27 = a 3 – 3 3 = (a – 3) [a 2 + (a x 3) + 3 2 ] [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] = (a – 3) (a 2 + 3a + 9)

2. 1 – 8a 3

Solution:

We have, 1 – 8a 3 = 1 3 – (2a) 3 = (1 – 2a) [1 2 + (1 x 2a) + (2a) 2 ] = (1 – 2a) (1 + 2a + 4a 2 ) [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )]

3. 64 – a 3 b 3

Solution:

We have, 64 – a 3 b 3 = 4 3 – (ab) 3 = (4 – ab) [4 2 + (4 x ab) + (ab) 2 ] = (4 – ab) (16 + 4ab + a 2 b 2 ) [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )]

4. a 6 + 27b 3

Solution:

We have, a 6 + 27b 3 = (a 2 ) 3 + (3b) 3 = (a 2 + 3b) [(a 2 ) 2 – (a 2 x 3b) + (3b) 2 ] = (a 2 + 3b) (a 4 – 3a 2 b + 9b 2 ) [As, a 3 + b 3 = (a + b) (a 2 – ab + b 2 )]

5. 3x 7 y – 81x 4 y 4

Solution:

We have, 3x 7 y – 81x 4 y 4 = 3xy (x 6 – 27x 3 y 3 ) = 3xy [(x 2 ) 3 – (3xy) 3 ] = 3xy (x 2 – 3xy) [(x 2 ) 2 + (x 2 × 3xy) + (3xy) 2 ] [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] = 3xy (x 2 – 3xy) (x 4 + 3x 3 y + 9x 2 y 2 ) = 3xy. x(x – 3y). x 2 (x 2 + 3xy + 9y 2 ) [Taking common from terms] = 3x 4 y (x – 3y) (x 2 + 3xy + 9y 2 )

6. a 3 – 27/a 3

Solution:

We have, a 3 – 27/a 3 = a 3 – (3/a) 3 = (a – 3/a) [a 2 + a x 3/a + (3/a) 2 ] [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] = (a – 3/a) (a 2 + 3 + 9/a 2 )

7. a 3 + 0.064

Solution:

We have, a 3 + 0.064 = a 3 + (0.4) 3 = (a + 0.4) [a 2 – (a x 0.4) + 0.4 2 ] = (a + 0.4) (a 2 – 0.4a + 0.16) [As, a 3 + b 3 = (a + b) (a 2 – ab + b 2 )]

8. a 4 – 343a

Solution:

We have, a 4 – 343a = a (a 3 – 343) = a (a 3 – 7 3 ) = a (a – 7) [a 2 + (a x 7) + 7 2 ] [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] = a (a – 7) (a 2 + 7a + 49)

9. (x – y) 3 – 8x 3

Solution:

We have, (x – y) 3 – 8x 3 = (x – y) 3 – (2x) 3 = (x – y – 2x) [(x – y) 2 + 2x(x – y) + (2x) 2 ] [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] = (-x – y) [x 2 + y 2 – 2xy + 2x 2 – 2xy + 4x 2 ] = -(x + y) [7x 2 – 4xy + y 2 ]

10. 8a 3 /27 – b 3 /8

Solution:

We have, 8a 3 /27 – b 3 /8 = (2a/3) 3 – (b/2) 3 = (2a/3 – b/2) [(2a/3) 2 + (2a/3 x b/2) + (b/2) 2 ] [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] = (2a/3 – b/2) (4a 2 /9 + ab/3 + b 2 /4)

11. a 6 – b 6

Solution:

We have, a 6 – b 6 = (a 3 ) 2 – (b 3 ) 2 = (a 3 + b 3 ) (a 3 – b 3 ) [As x 2 – y 2 = (x + y) (x – y)] Now, = [(a + b) (a 2 – ab + b 2 )] [(a – b) (a 2 + ab + b 2 )] [Using identities] = (a + b) (a – b) (a 2 – ab + b 2 ) (a 2 + ab + b 2 )

12. a 6 – 7a 3 – 8

Solution:

We have, a 6 – 7a 3 – 8 By splitting the middle term, = a 6 – 8a 3 + a 3 – 8 = a 3 (a 3 – 8) + 1(a 3 – 8) = (a 3 + 1) (a 3 – 8) We know that, a 3 – b 3 = (a – b) (a 2 + ab + b 2 ) … (1) a 3 + b 3 = (a + b) (a 2 – ab + b 2 ) … (2) Now, (a 3 + 1) (a 3 – 8) = [(a + 1) (a 2 – a + 1)] [(a – 2)(a 2 + 2a + 4)] … [Using (1) and (2)] = (a + 1) (a – 2) (a 2 + 2a + 4) (a 2 – a + 1)

13. a 3 – 27b 3 + 2a 2 b – 6ab 2

Solution:

We have, a 3 – 27b 3 + 2a 2 b – 6ab 2 = [a 3 – (3b) 3 ] + 2ab(a – 3b) = (a – 3b) (a 2 + 3ab + 9b 2 ) + 2ab(a – 3b) [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] Now, taking (a – 3b) as common = (a – 3b) [(a 2 + 3ab + 9b 2 ) + 2ab] = (a – 3b) (a 2 + 5ab + 9b 2 )

14. 8a 3 – b 3 – 4ax + 2bx

Solution:

We have, 8a 3 – b 3 – 4ax + 2bx = (2a) 3 – b 3 – 2x(2a – b) = (2a – b) [(2a) 2 – 2ab + b 2 ] – 2x(2a – b) Taking (2a – b) as common, = (2a – b) [(4a 2 + 2ab + b 2 ) – 2x] = (2a – b) (4a 2 + 2ab + b 2 – 2x)

15. a – b – a 3 + b 3

Solution:

We have, a – b – a 3 + b 3 = (a – b) – (a 3 – b 3 ) = (a – b) – [(a – b) (a 2 + ab + b 2 )] [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] Now, taking (a – b) as common = (a – b) [1 – (a 2 + ab + b 2 )] = (a – b) (1 – a 2 – ab – b 2 )

16. 2x 3 + 54y 3 – 4x – 12y

Solution:

We have, 2x 3 + 54y 3 – 4x – 12y = 2(x 3 + 27y 3 – 2x – 6y) Now, = 2 {[(x) 3 + (3y) 3 ] – 2(x + 3y)} = 2 {[(x + 3y) (x 2 – 3xy + 9y 2 )] – 2(x + 3y)} [As, a 3 + b 3 = (a + b) (a 2 – ab + b 2 )] = 2 (x + 3y) (x 2 – 3xy + 9y 2 – 2)

17. 1029 – 3x 3

Solution:

We have, 1029 – 3x 3 = 3(343 – x 3 ) = 3(7 3 – x 3 ) = 3(7 – x) (7 2 + 7x + x 2 ) [As, a 3 – b 3 = (a – b) (a 2 + ab + b 2 )] = 3(7 – x) (49 + 7x + x 2 )

18. Show that:

(i) 13 3 – 5 3 is divisible by 8

(ii)35 3 + 27 3 is divisible by 62

Solution:

(i) We have, (13 3 – 5 3 ) Now, using identity (a 3 – b 3 ) = (a – b) (a 2 + ab + b 2 ) = (13 – 5) (13 2 + 13 × 5 + 5 2 ) = 8 × (169 + 65 + 25) Hence, the number is divisible by 8. (ii) (35 3 + 27 3 ) Now, using identity (a 3 + b 3 ) = (a + b) (a 2 – ab + b 2 ) = (35 + 27) (35 2 + 35× 27 + 27 2 ) = 62 × (35 2 + 35 × 27 + 27 2 ) Hence, the number is divisible by 62.

19. Evaluate:

Selina Solutions Icse Class 9 Mathematics Chapter - Factorisation

Solution:

Let a = 5.67 and b = 4.33 Then, Selina Solutions Icse Class 9 Mathematics Chapter - Factorisation - 1 = a + b = 5.67 + 4.33 = 10

Exercise 5(E)

Factorize:

1. x 2 + 1/4x 2 + 1 – 7x – 7/2x

Solution:

We have, Concise Selina Solutions for Class 9 Maths Chapter 5 Factorisation - 1A = [x 2 + 1/(2x) 2 + 2 × x × 1/(2x)] – 7 [x + 1/(2x)] = (x + 1/2x) 2 – 7(x + 1/x) Taking out (x + 1/2x) as common, = (x + 1/2x) (x + 1/2x – 7)

2. 9a 2 + 1/9a 2 – 2 – 12a + 4/3a

Solution:

Concise Selina Solutions for Class 9 Maths Chapter 5 Factorisation - 1

3. x 2 + (a 2 + 1) x/a + 1

Solution:

Concise Selina Solutions for Class 9 Maths Chapter 5 Factorisation - 2

4. x 4 + y 4 – 27x 2 y 2

Solution:

We have, x 4 + y 4 – 27x 2 y 2 = x 4 + y 4 – 2x 2 y 2 – 25x 2 y 2 = [(x 2 ) + (y 2 ) – 2x 2 y 2 ] – 25x 2 y 2 = (x 2 – y 2 ) – (5xy) 2 = (x 2 – y 2 – 5xy) (x 2 – y 2 + 5xy) [As x 2 – y 2 = (x + y)(x – y)]

5. 4x 4 + 9y 4 + 11x 2 y 2

Solution:

We have, 4x 4 + 9y 4 + 11x 2 y 2 = 4x 4 + 9y 4 + 12x 2 y 2 – x 2 y 2 = (2x 2 ) 2 + (3y 2 ) 2 + 2(2x 2 )(3y 2 ) – (xy) 2 = (2x 2 + 3y 2 ) 2 – (xy) 2 = (2x 2 + 3y 2 – xy) (2x 2 + 3y 2 + xy) [As x 2 – y 2 = (x + y)(x – y)]

6. x 2 + 1/x 2 – 3

Solution:

We have, x 2 + 1/x 2 – 3 = x 2 + 1/x 2 – 2 – 1 = [x 2 + 1/x 2 – (2 × x × 1/x)] – 1 2 = (x – 1/x) 2 – 1 2 = (x – 1/x – 1) (x – 1/x + 1) [As x 2 – y 2 = (x + y)(x – y)]

7. a – b – 4a 2 + 4b 2

Solution:

We have, a – b – 4a 2 + 4b 2 = (a – b) – 4(a 2 – b 2 ) = (a – b) – 4(a – b)(a + b) [As x 2 – y 2 = (x + y)(x – y)] Taking (a – b) common, = (a – b) [1 – 4(a + b)] = (a – b) [1 – 4a – 4b]

8. (2a – 3) 2 – 2 (2a – 3) (a – 1) + (a – 1) 2

Solution:

We have, (2a – 3) 2 – 2(2a – 3)(a – 1) + (a – 1) 2 Comparing with the identity, (a – b) 2 = a 2 – 2ab + b 2 = [(2a – 3) – (a – 1)] 2 = (2a – a – 3 + 1) 2 = (a – 2) 2

9. (a 2 – 3a) (a 2 – 3a + 7) + 10

Solution:

Let’s substitute (a 2 – 3a) = x Then the given expression becomes, = x(x + 7) + 10 = x 2 + 7x + 10 = x 2 + 5x + 2x + 10 [By splitting the middle term] = x(x + 5) + 2(x + 5) = (x + 2) (x + 5) Resubstituting the value of x, = (a 2 – 3a + 2) (a 2 – 3a + 5) = (a 2 – 3a + 5) (a 2 – 3a + 2) = (a 2 – 3a + 5) [a 2 – 2a – a + 2] [By splitting the middle term] = (a 2 – 3a + 5) [a(a – 2) – 1(a – 5)] = (a 2 – 3a + 5) [(a – 1) (a – 2)] = (a 2 – 3a + 5) (a – 1) (a – 2)

10. (a 2 – a) (4a 2 – 4a – 5) – 6

Solution:

Let’s a 2 – a = x Then the expression becomes, = x(4x – 5) – 6 = 4x 2 – 5x – 6 = 4x 2 – 8x + 3x – 6 = 4x(x – 2) + 3(x – 2) = (4x + 3) (x – 2) Resubstituting the value of x, = (4a 2 – 4a + 3) (a 2 – a – 2) = (4a 2 – 4a + 3) (a 2 – 2a + a – 2) = (4a 2 – 4a + 3) [a(a – 2) + 1(a – 2)] = (4a 2 – 4a + 3) [(a + 1) (a – 2)] = (4a 2 – 4a + 3) (a + 1) (a – 2)

11. x 4 + y 4 – 3x 2 y 2

Solution:

We have, x 4 + y 4 – 3x 2 y 2 = (x 4 + y 4 – 2x 2 y 2 ) – x 2 y 2 = (x 2 – y 2 ) – (xy) 2 [As x 2 – y 2 = (x + y)(x – y)] = (x 2 – y 2 – xy) (x 2 – y 2 + xy)

12. 5a 2 – b 2 – 4ab + 7a – 7b

Solution:

We have, 5a 2 – b 2 – 4ab + 7a – 7b = 4a 2 + a 2 – b 2 – 4ab + 7a – 7b = a 2 – b 2 + 4a 2 – 4ab + 7a – 7b = (a 2 – b 2 ) + 4a(a – b) + 7(a – b) = (a + b)(a – b) + 4a(a – b) + 7(a – b) [As x 2 – y 2 = (x + y)(x – y)] = (a – b) [(a + b) + 4a + 7] = (a – b) (5a + b + 7)

13. 12(3x – 2y) 2 – 3x + 2y – 1

Solution:

We have, 12(3x – 2y) 2 – 3x + 2y – 1 = 12(3x – 2y) 2 – (3x – 2y) – 1 Let’s substitute (3x – 2y) = a Then, the expression becomes = 12a 2 – a – 1 = 12a 2 – 4a + 3a – 1 = 4a (3a – 1) + 1(3a – 1) = (4a + 1) (3a – 1) Now, resubstituting the value of ‘a’ in the above = [4(3x – 2y) + 1] [3(3x – 2y) – 1] = (12x – 8y + 1) (9x – 6y – 1)

14. 4(2x – 3y) 2 – 8x+12y – 3

Solution:

We have, 4(2x – 3y) 2 – 8x+12y – 3 = 4(2x – 3y) 2 – 4(2x + 3y) – 3 Let’s substitute (2x – 3y) = a = 4(a 2 ) – 4a – 3 = 4a 2 – 6a + 2a – 3 [By splitting the middle term] = 2a(2a – 3) + 1(2a – 3) = (2a – 3) (2a + 1) Now, resubstituting the value of ‘a’ in the above = [2(2x – 3y) – 3] [2(2x – 3y) + 1] = (4x – 6y – 3) (4x – 6y + 1)

15. 3 – 5x + 5y – 12(x – y) 2

Solution:

We have, 3 – 5x + 5y – 12(x – y) 2 = 3 – 5(x – y) – 12(x – y) 2 Let’s substitute (x – y) = a = 3 – 5a – 12a 2 = 3 – 9a + 4a – 12a 2 [By splitting the middle term] = 3(1 – 3a) + 4a(1 – 3a) = (1 – 3a) (4a + 3) Now, resubstituting the value of ‘a’ in the above = [1 – 3(x – y)] [4(x – y) + 3] = (1 – 3x + 3y) (4x – 4y + 3)

16. 9x 2 + 3x – 8y – 64y 2

Solution:

We have, 9x 2 + 3x – 8y – 64y 2 On rearranging, = 9x 2 – 64y 2 + 3x – 8y = [(3x) 2 – (8y) 2 ] + (3x – 8y) = (3x – 8y) (3x + 8y) + (3x – 8y) [As x 2 – y 2 = (x + y)(x – y)] Taking (3x – 8y) as common, = (3x – 8y) (3x + 8y + 1)

17. 2√3x 2 + x – 5√3

Solution:

We have, 2√3x 2 + x – 5√3 By splitting the middle term, = 2√3x 2 + 6x – 5x – 5√3 = 2√3x(x + √3) – 5(x + √3) = (2√3x – 5) (x + √3)

18. ¼ (a + b) 2 – 9/16 (2a – b) 2

Solution:

Concise Selina Solutions for Class 9 Maths Chapter 5 Factorisation - 3

19. 2(ab + cd) – a 2 – b 2 + c 2 + d 2

Solution:

We have, 2(ab + cd) – a 2 – b 2 + c 2 + d 2 = 2ab + 2cd – a 2 – b 2 + c 2 + d 2 On rearranging and grouping, we get = (c 2 + d 2 + 2cd) – (a 2 + b 2 – 2ab) = (c + d) 2 – (a – b) 2 = [c + d – (a – b)] [c + d + (a – b)] [As x 2 – y 2 = (x + y)(x – y)] = (c + d – a + b) (c + d + a – b)

20. Find the value of:

(i) 987 2 – 13 2

(ii) (67.8) 2 – (32.2) 2

(iii) [(6.7) 2 – (3.3) 2 ]/ (6.7 – 3.3)

(iv) [(18.5) 2 – (6.5) 2 ]/ (18.5 – 6.5)

Solution:

(i) We have, 987 2 – 13 2 = (987 + 13) (987 – 13) = 1000 x 974 = 974000 (ii) We have, (67.8) 2 – (32.2) 2 = (67.8 + 32.2) (67.8 – 32.2) = 100 x 35.6 = 3560 Concise Selina Solutions for Class 9 Maths Chapter 5 Factorisation - 4

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ICSE Class 9 Maths Selina Solutions Chapter wise
Chapter 1 Rational and Irrational Numbers Chapter 11 Inequalities
Chapter 2 Compound Interest (Without Using Formula) Chapter 12 Mid-Point and Its Converse (Including Intercept Theorem)
Chapter 3 Compound Interest (Using Formula) Chapter 13  Pythagoras Theorem
Chapter 4 Expansions Chapter 14 Rectilinear Figures (Quadrilaterals: Parallelogram, Rectangles, Rhombus, Square, and Trapezium)
Chapter 5 Factorisation Chapter 15 Construction of Polygons
Chapter 6 Simultaneous (Linear) Equations (Including Problems) Chapter 16 Area Theorems
Chapter 7 Indices (Exponents) Chapter 17 Circle
Chapter 8 Logarithms Chapter 18 Statistics
Chapter 9 Triangles (Congruency in Triangles) Chapter 19 Mean and Median
Chapter 10 Isosceles Triangles Chapter 20 Area and Perimeter of Plane Figures

ICSE Class 9 Maths Selina Solutions Chapter 5 FAQs

Q1. What is Factorization?

Ans. In mathematics, Factorization is writing a number or other mathematical object as the product of several factors, usually the product of smaller or simpler similar objects. For example, 3 × 5 is an integer factorization of 15 and is a polynomial factorization of x²- 4.

Q2. Where can I get the ICSE Class 9 Maths Selina Solutions Chapter 5 Factorization?

Ans. You can get the ICSE Class 9 Maths Selina Solutions Chapter 5 - Factorization at PW blog section.

Q3. Is ICSE Class 9 Maths Selina Solutions Chapter 5 PDF available to download?

Ans. Yes, ICSE Class 9 Maths Selina Solutions Chapter 5 PDF is available for download on our page.

Q4. What are the Benefits of ICSE Class 9 Maths Selina Solutions Chapter 5 Factorization?

Ans. The details Benefits of ICSE Class 9 Maths Selina Solutions Chapter 5 Factorization are provided in the above article.

Q5. Are ICSE Class 9 Maths Selina Solutions Chapter 5 useful for exam preparation?

Ans. Yes, these ICSE Class 9 Maths Selina Solutions Chapter 5 are extremely beneficial for exam preparation. They cover all topics and exercises prescribed in the ICSE curriculum, ensuring comprehensive revision and practice.
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