RS Aggarwal Solutions for Class 8 Maths Chapter 7- Factorisation, are provided here. Our expert faculty team has prepared solutions to help you with your exam preparation to acquire good marks in Maths. RS Aggarwal Solutions for Class 8 MathsÂ comes in very handy at this point. Our solution module utilises various shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language. If you wish to secure an excellent score, solving RS Aggarwal Class 8 Solutions is an utmost necessity. These solutions will help you in gaining knowledge and strong command over the subject. Practising the textbook questions will help you in analysing your level of preparation and understanding of the concept.

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## Exercise 7A

**Factorise:**

1.iÂ **12x + 15**

Solution:

By taking 3 as a common factor for the above equation we get,

12x + 15 = 3(4x + 5)

**ii 14m â€“ 21**

Solution:

By taking 7 as a common factor for the above equation we get,

14m – 21 = 7(2m – 3)

**iii 9n-12n ^{2}**

Solution:

By taking 3n as a common factor for the above equation we get,

9n â€“ 12n^{2} = 3(3 â€“ 4n)

2.

**i 16a ^{2} â€“ 24ab**

Solution:

letâ€™s take HCF of above equation

By taking 8a as a common factor for the above equation we get,

16a^{2} â€“ 24ab = 8a (2a â€“ 3b)

**ii 15ab ^{2} â€“ 20a^{2}b**

Solution:

letâ€™s take HCF of above equation

By taking 5ab as a common factor for the above equation we get,

15ab^{2} â€“ 20a^{2}b = 5ab (3b â€“ 4a)

**iii 12x ^{2}y^{3} â€“ 21x^{3}y^{2}**

Solution:

letâ€™s take HCF of above equation

By taking 3x^{2}y^{2} as a common factor for the above equation we get,

12x^{2}y^{3} â€“ 21x^{3}y^{2 }= 3x^{2}y^{2 }(4y â€“ 7x)

3.

**i 24x ^{3} â€“ 36x^{2}y**

Solution:

letâ€™s take HCF of above equation

By taking 12x^{2} as a common factor for the above equation we get,

24x^{3} â€“ 36x^{2}y = 12x^{2} (2x â€“ 3y)

**ii 10x ^{3} -15x^{2}**

Solution:

letâ€™s take HCF of above equation

By taking 5x^{2} as a common factor for the above equation we get,

10x^{3} -15x^{2 }= 5x^{2} (2x – 3)

**iii 36x ^{3}y â€“ 60x^{2}y^{3}z**

Solution:

letâ€™s take HCF of above equation

By taking 12x^{2}y as a common factor for the above equation we get,

36x^{3}y â€“ 60x^{2}y^{3}z = 12x^{2}y (3x â€“ 5y^{2}z)

4.

**i 9x ^{3} â€“ 6x^{2} + 12x**

Solution:

letâ€™s take HCF of above equation

By taking 3x as a common factor for the above equation we get,

9x^{3} â€“ 6x^{2} + 12x = 3x (3x^{2} â€“ 2x + 4)

**ii 8x ^{2} â€“ 72xy + 12x**

Solution:

letâ€™s take HCF of above equation

By taking 4x as a common factor for the above equation we get,

8x^{2} â€“ 72xy + 12x = 4x (2x -18y +3)

**iii 18a ^{3}b^{3} â€“ 27a^{2}b^{3} + 36a^{3}b^{2}**

Solution:

letâ€™s take HCF of above equation

By taking 9a^{2}b^{2} as a common factor for the above equation we get,

18a^{3}b^{3} â€“ 27a^{2}b^{3} + 36a^{3}b^{2 }= 9a^{2}b^{2} (2ab â€“ 3b + 4a)

5.

**i 14x ^{3} + 21x^{4}y -28x^{2}y^{2}**

Solution:

letâ€™s take HCF of above equation

By taking 7x^{2} as a common factor for the above equation we get,

14x^{3} + 21x^{4}y -28x^{2}y^{2} = 7x^{2} (2x + 3x^{2}y â€“ 4y^{2})

**ii -5 -10t + 20t ^{2}**

Solution:

letâ€™s take HCF of above equation

By taking 5 as a common factor for the above equation we get,

-5 -10t + 20t^{2 }= -5 (1 +2t -4t^{2})

6.

**i x(x+3) + 5(x+3)**

Solution:

By taking x+3 as a common factor for the above equation we get,

x (x+3) + 5(x+3) = (x+3) (x + 5)

**ii 5x(x-4) -7(x-4)**

Solution:

By taking x-4 as a common factor for the above equation we get,

5x(x-4) – 7(x-4) = (x-4) (5x – 7)

**iii 2m(1-n) + 3(1-n)**

Solution:

By taking 1-n as a common factor for the above equation we get,

2m (1-n) + 3(1-n) = (1-n) (2m + 3)

**7. 6a(a-2b) + 5b(a-2b)**

Solution: By taking a-2b as a common factor for the above equation we get,

6a (a-2b) + 5b (a-2b) = (a-2b) (6a + 5b)

**8. x ^{3}(2a-b) + x^{2}(2a-b)**

Solution: By taking 2a-b as a common factor for the above equation we get,

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

**9. 9a(3a-5b) â€“ 12a ^{2}(3a-5b)**

Solution: By taking 3a-5b as a common factor for the above equation we get,

9a (3a-5b) â€“ 12a^{2} (3a-5b) = (3a-5b) (9a â€“ 12a^{2})

**10. (x+5) ^{2} â€“ 4(x+5)**

Solution: By taking (x+5) as a common factor for the above equation we get,

(x+5)^{2} â€“ 4(x+5) = (x+5) ((x+5) – 4)

= (x+5) (x + 5 – 4)

= (x+5) (x+1)

**11. 3(a-2b) ^{2} â€“ 5(a-2b)**

Solution: By taking (a-2b) as a common factor for the above equation we get,

3(a-2b)^{ 2} â€“ 5(a-2b) = (a-2b) (3(a-2b) – 5)

= (a-2b) (3a -6b -5)

**12. 2a +6b -3(a+3b) ^{2} **

Solution: 2(a+3b) â€“ 3(a+3b)^{ 2}

Now by taking (a+3b) as a common factor for the above equation we get,

2(a+3b) â€“ 3(a+3b)^{ 2 }= (a+3b) (2 â€“ 3(a+3b))

= (a+3b) (2- 3a -9b)

**13. 16(2p-3q) ^{2} â€“ 4(2p-3q)**

Solution: By taking (2p-3q) as a common factor for the above equation we get,

16(2p-3q)^{ 2} â€“ 4(2p-3q) = (2p-3q) (16(2p-3q) – 4)

= (2p-3q) (32p â€“ 48q -4)

= (2p-3q) 4(8p -12q -1)

**14. x(a-3) + y(3-a)**

Solution: x(a-3) â€“y(a-3)

By taking (a-3) as a common factor for the above equation we get,

x(a-3) â€“y(a-3) = (a-3) (x-y)

## Exercise 7B

**Factorise:**

**1. x ^{2} â€“ 36**

Solution:

x^{2} â€“ 36 can be written as (x)^{ 2} â€“ (6)^{2}

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

(x)^{ 2} â€“ (6)^{2} = (x+6) (x-6)

**2. 4a ^{2} â€“ 9**

Solution:

4a^{2} â€“ 9 can be written as (2a)^{ 2} â€“ (3)^{2}

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

(2a)^{ 2} â€“ (3)^{2} = (2a+3) (2a-3)

**3. 81 â€“ 49x ^{2}**

Solution:

81 â€“ 49x^{2} can be written as (9)^{2} â€“ (7x)^{ 2}

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

(9)^{2} â€“ (7x)^{ 2} = (9+7x) (9-7x)

**4. 4x ^{2} â€“ 9y^{2}**

Solution:

4x^{2} â€“ 9y^{2} can be written as (2x)^{ 2} â€“ (3y)^{ 2}

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

(2x)^{ 2} â€“ (3y)^{ 2} = (2x+3y) (2x-3y)

**5. 16a ^{2} â€“ 225b^{2}**

Solution:

16a^{2} â€“ 225b^{2} can be written as (4a)^{ 2} â€“ (15b)^{ 2}

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

(4a)^{ 2} â€“ (15b)^{ 2} = (4a+15b) (4a-15b)

**6. 9a ^{2}b^{2} â€“ 25**

Solution:

9a^{2}b^{2} â€“ 25 can be written as (3ab)^{ 2} â€“ (5)^{2}

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

(3ab)^{ 2} â€“ (5)^{2} = (3ab+5) (3ab-5)

**7. 16a ^{2} -144**

Solution:

16a^{2} -144 can be written as (4a)^{ 2} â€“ (12)^{2}

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

(4a)^{ 2} â€“ (12)^{2 }= (4a+12) (4a-12)= 4(a+3) 4(a-3)= 16(a+3) (a-3)

**8. 63a ^{2} -112b^{2}**

Solution:

By simplifying further the above equation is 7(9a^{2} â€“ 16b^{2})

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

7(9a^{2} â€“ 16b^{2}) = 7((3a)^{ 2} â€“ (4b)^{ 2})= 7(3a+4b) (3a-4b)

**9. 20a ^{2} â€“ 45b^{2}**

Solution:

By simplifying further the above equation is 5(4a^{2} â€“ 9b^{2})

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

5(4a^{2} â€“ 9b^{2}) = 5((2a)^{ 2} â€“ (3b)^{ 2})= 5(2a+3b) (2a-3b)

**10. 12x ^{2} -27**

Solution:

By simplifying further the above equation is 3(4x^{2} â€“ 9)

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

3(4x^{2} â€“ 9) = 3((2x)^{ 2} â€“ (3)^{2})= 3(2x+3) (2x-3)

**11. x ^{3} – 64x**

Solution:

x^{3} â€“ 64x can be written as x(x^{2} â€“ 64)

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

x (x^{2} â€“ 64) = x ((x)^{2} â€“ (8)^{2})= x(x+8) (x-8)

**12. 16x ^{5} -144x^{3}**

Solution:

16x^{5} -144x^{3} can be written as 16x^{3}(x^{2} â€“ 9)

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

16x^{3}(x^{2} â€“ 9) = 16x^{3}((x)^{ 2} â€“ (3)^{2})=16x^{3}(x+3) (x-3)

**13. 3x ^{5} â€“ 48x^{3}**

Solution:

3x^{5} â€“ 48x^{3} can be written as 3x^{3}(x^{2} â€“ 16)

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

3x^{3}(x^{2} â€“ 16) = 3x^{3}((x)^{ 2} â€“ (4)^{2})= 3x^{3}(x+4) (x-4)

**14. 16p ^{3} -4p**

Solution:

16p^{3} -4p can be written as 4p (4p^{2} â€“ 1)

By using the formula a^{2} â€“ b^{2} = (a+b) (a-b)

Now solving for the above equation

4p (4p^{2} â€“ 1) = 4p ((2p)^{ 2} â€“ (1)^{2})= 4p (2p+1) (2p-1)

## Exercise 7C

**Factorise:**

**1. x ^{2} + 8x + 16**

Solution:

we solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, x^{2} + (4)^{2} + 2 (x) (4) = (x+4)^{2}

**2. x ^{2} + 14x + 49**

Solution:

we solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, x^{2} + (7)^{2} + 2 (x) (7) = (x+7)^{2}

**3. 1 + 2x + x ^{2}**

Solution:

we rewrite it as x^{2} + 2x + 1

We solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, x^{2} + (1)^{2} + 2 (x) (1) = (x+1)^{2} = (x+1) (x+1)

**4. 9 + 6z + z ^{2}**

Solution:

we rewrite it as z^{2} + 6z + 9

We solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, z^{2} + (3)^{2} + 2 (z) (3) = (z+3)^{2}

**5. x ^{2} + 6ax + 9a^{2}**

Solution:

We solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, x^{2} + (3)^{2} + 2 (x) (3a) = (x+3a)^{ 2}

**6. 4y ^{2} +20y +25**

Solution:

We solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, (2y)^{ 2} + (5)^{2} + 2 (2y) (5) = (2y+5)^{ 2}

**7. 36a ^{2} + 36a +9**

Solution:

We solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, (6a)^{ 2} + (3)^{2} + 2 (6a) (3) = (6a+3)^{ 2}

**8. 9m ^{2} + 24m + 16**

Solution:

We solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, (3m)^{ 2} + (4)^{2} + 2 (3a) (4) = (3m+4)^{ 2}

**9. z ^{2} + z + Â¼**

Solution:

We solve by using the formula (a+b)^{ 2} = a^{2} + b^{2} + 2ab

We get, (z)^{ 2} + (1/2)^{2} + 2 (z) (1/2) = (z+1/2)^{ 2}

## Exercise 7D

**Factorise:**

**1. x ^{2} +5x + 6**

Solution:

First find the two numbers whose sum=5 and product= 6

Clearly, the numbers are 2 and 3

âˆ´ we get, x^{2} +5x + 6 = x^{2 }+ 2x +3x + 6= x(x+2) + 3(x+2)= (x+3) (x+2)

**2. y ^{2} +10y +24**

Solution:

First find the two numbers whose sum=10 and product= 24

Clearly, the numbers are 6 and 4

âˆ´ we get, y^{2} +10y + 24 = y^{2 }+ 6y + 4y + 24= y(y+6) + 4(y+6)= (y+4) (y+6)

**3. z ^{2} +12z +27**

Solution:

First find the two numbers whose sum=12 and product= 27

Clearly, the numbers are 9 and 3

âˆ´ we get, z^{2} +12z + 27 = z^{2 }+ 9z + 3z + 27= z (z+9) + 3(z+9)= (z+3) (z+9)

**4. p ^{2} + 6p + 8**

Solution:

First find the two numbers whose sum=6 and product= 8

Clearly, the numbers are 4 and 2

âˆ´ we get, p^{2} +6p + 8 = p^{2 }+ 4p + 2p + 8= p (p+4) + 2(p+4)= (p+2) (p+4)

**5. x ^{2} +15x + 56**

Solution:

First find the two numbers whose sum=15 and product=56

Clearly, the numbers are 7 and 8

âˆ´ we get, x^{2} +15x + 56 = x^{2 }+ 7x +8x + 56= x(x+7) + 8(x+7)= (x+7) (x+8)

**6. y ^{2} +19y +60**

Solution:

First find the two numbers whose sum=19 and product= 60

Clearly, the numbers are 15 and 4

âˆ´ we get, y^{2} +19y + 60 = y^{2 }+ 15y + 4y + 60= y(y+15) + 4(y+15)= (y+15) (y+4)

**7. x ^{2} +13x + 40**

Solution:

First find the two numbers whose sum=13 and product=40

Clearly, the numbers are 8 and 5

âˆ´ we get, x^{2} +13x + 40 = x^{2 }+ 8x +5x + 40= x(x+8) + 5(x+8)= (x+8) (x+5)

**8. q ^{2} -10q + 21**

Solution:

First find the two numbers whose sum= -10 and product=21

Clearly, the numbers are 7 and 3

âˆ´ we get, q^{2} -10q + 21 = q^{2 }â€“ 7q -3q + 21= q (q-7) – 3(q-7)= (q-7) (q-3)

**9. p ^{2} + 6p -16**

Solution:

First find the two numbers whose sum=6 and product= -16

Clearly, the numbers are 8 and 2

âˆ´ we get, p^{2} +6p – 16 = p^{2 }+8p – 2p -16= p (p+8) – 2(p+8)= (p+8) (p-2)

**10. x ^{2} – 10x + 24**

Solution:

First find the two numbers whose sum= -10 and product=24

Clearly, the numbers are 6 and 4

âˆ´ we get, x^{2} – 10x + 24 = x^{2 }– 6x -4x + 24= x(x-6) – 4(x-6)= (x-6) (x-4)

**11. x ^{2} – 23x + 42**

Solution:

First find the two numbers whose sum= -23 and product=42

Clearly, the numbers are 21 and 2

âˆ´ we get, x^{2} – 23x + 42 = x^{2 }– 21x – 2x + 42= x(x-21) – 2(x-21)= (x-21) (x-2)

**12. x ^{2} – 17x + 16**

Solution:

First find the two numbers whose sum= -17 and product=16

Clearly, the numbers are 16 and 1

âˆ´ we get, x^{2} – 17x + 16 = x^{2 }– 16x – 1x + 16= x(x-16) – 1(x-16)= (x-16) (x-1)

**13. y ^{2} – 21y +90**

Solution:

First find the two numbers whose sum= -21 and product= 90

Clearly, the numbers are 15 and 6

âˆ´ we get, y^{2} – 21y + 90 = y^{2 }– 15y – 6y + 90= y(y-15) – 6(y-15)= (y-15) (y-6)

**14. x ^{2} – 22x + 117**

Solution:

First find the two numbers whose sum= -22 and product=117

Clearly, the numbers are 13 and 9

âˆ´ we get, x^{2} – 22x + 117 = x^{2 }– 13x – 9x + 117= x(x-13) – 9(x-13)= (x-13) (x-9)

**15. x ^{2} – 9x + 20**

Solution:

First find the two numbers whose sum= -9 and product=20

Clearly, the numbers are 5 and 4

âˆ´ we get, x^{2} – 9x + 20 = x^{2 }– 5x – 4x + 20= x(x-5) – 4(x-5)= (x-5) (x-4)

**16. x ^{2} + x – 132**

Solution:

First find the two numbers whose sum= 1 and product= -132

Clearly, the numbers are 12 and 11

âˆ´ we get, x^{2} + x – 132 = x^{2 }+ 12x – 11x – 132= x(x+12) – 11(x+12)= (x+12) (x-11)

**17. x ^{2} + 5x – 104**

Solution:

First find the two numbers whose sum= 5 and product= -104

Clearly, the numbers are 13 and 8

âˆ´ we get, x^{2} + 5x – 104 = x^{2 }+ 13x – 8x – 104= x(x+13) – 8(x+13)= (x+13) (x-8)

**18. y ^{2} + 7y – 144**

Solution:

First find the two numbers whose sum= 7 and product= -144

Clearly, the numbers are 16 and -9

âˆ´ we get, y^{2} + 7y – 144 = y^{2 }+ 16y – 9y – 144= y(y+16) – 9(y+16)= (y+16) (y-9)

**19. z ^{2} +19z -150**

Solution:

First find the two numbers whose sum=19 and product= -150

Clearly, the numbers are 25 and 6

âˆ´ we get, z^{2} +19z – 150 = z^{2 }+ 25z – 6z – 150= z(z+25) – 6(z+25)= (z+25) (z-6)

**20. y ^{2} + y – 72**

Solution:

First find the two numbers whose sum= 1 and product= -72

Clearly, the numbers are 9 and 8

âˆ´ we get, y^{2} + y – 72 = y^{2 }+ 9y – 8y – 72= y(y+9) – 8(y+9)= (y+9) (y-8)

**21. a ^{2} + 6a – 91**

Solution:

First find the two numbers whose sum= 6 and product= -91

Clearly, the numbers are 13 and 7

âˆ´ we get, a^{2} + 6a – 91 = a^{2 }+ 13a â€“ 7a – 91= a (a+13) – 7(a+13)= (a+13) (a-7)

## Exercise 7E

**Select the correct answer in each of the following:**

**1. (7a ^{2} â€“ 63b^{2}) = ?**

**(7a â€“ 9b) (9a + 7b)****(7a â€“ 9b) (7a + 9b)****9(a â€“ 3b) (a + 3b)****7(a â€“ 3b) (a + 3b)**

Solution:

let us consider (7a^{2} â€“ 63b^{2}) = 7(a^{2} â€“ 9b^{2}) by taking 7 as the common factor

By using the formula a^{2} – b^{2} = (a â€“ b) (a + b)** = **7 (a â€“ 3b) (a + 3b)

**2. (2x â€“ 32x ^{3}) = ?**

**2(x-4)(x+4)****2x(1-2x)**^{2}**2x(1+2x)**^{2}**2x(1-4x)(1+4x)**

Solution:

let us consider (2x â€“ 32x^{3}) = 2x (1 â€“ 16x^{2}) by taking 2x as the common factor

By using the formula a^{2} – b^{2} = (a â€“ b) (a + b)** = **2x (1 â€“ 4x) (1 + 4x)

**3. x ^{3} -144x = ?**

**x(x-12)**^{2}**x(x+12)**^{2}**x(x-12)(x+12)****none of these**

Solution:

let us consider (x^{3} -144x) = x (x^{2} â€“ 144) by taking x as the common factor

By using the formula a^{2} – b^{2} = (a â€“ b) (a + b)** = **x (x â€“ 12) (x + 12)

**4. (2 â€“ 50x ^{2}) = ?**

**2(1-5x)**^{2}**2(1+5x)**^{2}**(2-5x)(2+5x)****2(1-5x)(1+5x)**

Solution: let us consider (2 -50x^{2}) = 2 (1 â€“ 25x^{2}) by taking 2 as the common factor

By using the formula a^{2} – b^{2} = (a â€“ b) (a + b)

** = **2 (1 â€“ 5x) (1 + 5x)

**5. a ^{2} + bc + ab + ac = ?**

**(a+b)(a+c)****(a+b)(b+c)****(b+c)(c+a)****a(a+b+c)**

Solution:

let us rearrange the terms and take a and c as a the common factor we get,

a^{2} + bc + ab + ac = a^{2} + ab + bc + ac = a (a+b) + c(a+b)= (a+b) (a+c)

**6. pq ^{2} + q(p-1) -1 =?**

**(pq+1)(q-1)****p(q+1)(q-1)****q(p-1)(q+1)****(pq-1)(q+1)**

Solution:

firstly lets simplify pq^{2} + q(p-1) -1 = pq^{2} + qp â€“q -1

Now take the common factor as pq we get,

pq^{2} + qp â€“q -1 = pq(q+1) -1(q+1)= (q+1) (pq-1)

**7. ab â€“mn +an â€“bm = ?**

**(a-b) (m-n)****(a-m) (b+n)****(a-n) (m+b)****(m-a) (n-b)**

Solution:

let us rearrange the terms and take a and m as a the common factor we get,

ab â€“mn +an â€“bm = ab + an â€“mn â€“ bm= a(b+n) â€“m(b+n)= (b+n) (a-m)

**8. ab â€“a â€“b +1 = ?**

**(a-1) (b-1)****(1-a) (1-b)****(a-1) (1-b)****(1-a) (b-1)**

Solution:

By taking a and -1 as the common factor we get,

ab â€“a â€“b +1 = a(b-1) -1(b-1)= (b-1) (a-1)

**9. x ^{2} â€“xz + xy â€“yz =?**

**(x-y)(x+z)****(x-y)(x-z)****(x+y)(x-z)****(x-y)(z-x)**

Solution:

By taking x and y as the common factor we get,

x^{2} â€“xz + xy â€“yz = x(x-z) + y(x-z)= (x-z) (x+y)

**10. (12m ^{2} -27) = ?**

**(2m-3)(3m-9)****3(2m-9)(3m-1)****3(2m-3)(2m+3)****None of these**

Solution:

By taking 3 as the common factor we get,

By using the formula a^{2} – b^{2} = (a â€“ b) (a + b)

(12m^{2} -27) = 3(4m^{2} -9)= 3(2m -3) (2m + 3)

## RS Aggarwal Solutions for Class 8 Maths Chapter 7 – Factorisation

Chapter 7- Factorisation contains 5 exercises and the RS Aggarwal solutions present on this page provide the solutions for the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

- Factorisation when a common monomial factor occurs in each term.
- Factorisation when a binomial is common.
- Factorisation by grouping.
- Factorisation when a given expression is the difference of two squares.
- Factorisation when a given expression is a perfect square.
- Factorisation of quadratic trinomials.

### Also, Access RS Aggarwal Solutions for Class 8 Maths Chapter 7 Factorisation Exercises

### Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 7 – Factorisation

RS Aggarwal Solutions for Class 8 Maths Chapter 7 – Factorisation, ensures that the students are thorough and familiar with the concepts. Regular revision of important concepts and formulas over time to time is the best way to strengthen your concepts. The solutions are designed in such a way that they are easy to understand and solve.

As the Chapter is about factors, factoring is a useful skill required in real life. Common applications include dividing something into equal halves, exchanging money, comparing prices, understanding time and making calculations during travel. In the RS Aggarwal Solutions book, many such exercise problems are given, which enhances familiarity with the concepts.