ML Aggarwal Factorisation Exe-11.1 Class 8 ICSE Ch-11 Maths Solutions. We Provide Step by Step Answer of  Exe-11.1 Questions for Factorisation as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

## ML Aggarwal Factorisation Exe-11.1 Class 8 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 8th Chapter-11 Factorisation Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-11.1 Questions Edition 2023-2024

### Factorisation Exe-11.1

ML Aggarwal Class 8 ICSE Maths Solutions

Page-193

#### Question 1.

(i) 8xy3 + 12x2y2
(ii) 15ax3 – 9ax2

(i) 8xy3 + 12x2y2 = 4xy2 (2y + 3x)
(ii) 15ax3 – 9ax2 = 3ax2 (5x – 3)

#### Question 2.

(i) 21 py2 – 56py
(ii) 4x3 – 6x2

(i) 21 py2 – 56py = 7py (3y – 8)
(ii) 4x3 – 6x2 = 2x2 (2x – 3)

#### Question 3.

(i) 25abc2 – 15a2b2c
(ii) x2yz + xy2z + xyz2

(i) 25abc2 – 15a2b2c = 5abc (5c – 3ab)
(ii) x2yz + xy2z + xyz2 = xyz(x + y + z)

#### Question 4.

(i) 8x3 – 6x2 + 10x
(ii) 14mn + 22m – 62p

(i) 8x3 – 6x2 + 10x = 2x (4x2 – 3x + 5)
(ii) 14mn + 22m – 62p = 2 (7mn + 11m – 31p)

#### Question 5.

(i) 18p2q2 – 24pq2 + 30p2q
(ii) 27a3b3 – 18a2b3 + 75a3b2

(i) 18p2q2 – 24 pq2 + 30p2q
= 6pq (3pq -4q + 5p)
(ii) 27a3b3 – 18a2b3 + 75a3b2
= 3a2b2 (9ab – 6b + 25a)

#### Question 6.

(i) 15a (2p – 3p) – 106 (2p – 3q)
(ii) 3a (x2 + y2) + 6b (x2 + y2)

(i) 15a (2p – 3q) – 10b (2p – 3q)
= (2p – 3q)(15a – 10b)
= (2p – 3q) (5) (3a – 2b)
= 5 (2p- 3q) (3a – 2b)

(ii) 3a (x2 + y2) + 66 (x2 + y2)
= (x2 + y2) (3a + 6b)
= (x2 + y2) (3) (a + 2b)
= 3 (x2 + y2) (a + 2b)

#### Question 7.

(i) 6(x + 2y)3 + 8(x + 2y)2
(ii) 14(a – 3b)3 – 21p(a – 3b)

(i) 6(x + 2y)3 + 8(x + 2y)2
(x + 2y)2 [6 (x + 2y) + 8] = (x + 2y)2 [6x + 12y + 8] = (x + 2y)2 (2) (3x + 6y + 4)
= 2 (x + 2y)2 (3x + 6y + 4)

(ii) 14(a – 3b)3 – 21 p(a – 3b)
= 7 [2 (a – 3b)3 -3p(a- 3b)] = 7 [(a – 3b) {2 (a – 3b)2 – 3p}] = 7 (a – 3b) [2 (a – 3b)2 – 3p]

#### Question 8. 10a (2p + q)3 – 15b (2p + q)2 + 35(2p + q)

10a (2p + q)3 – 15b (2p + q)2 + 35(2p + q)
= 5 [2a (2p + q)]3 – 3b (2p + q)2 + 7 (2p + q)
= 5(2p + q) [2a (2p + q)2 – 3b(2p + q) + 7]

— End of Factorisation Exe-11.1 Class 8 ICSE Maths Solutions :–