ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.1

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.1

Factorise the following (1 to 8) polynomials:
Question 1.
(i) 8xy³ + 12x²y²
(ii) 15ax³ – 9ax²
Solution:
(i) 8xy³ + 12x²y² = 4xy² (2y + 3x)
(ii) 15ax³ – 9ax² = 3ax² (5x – 3)

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

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

Question 4.
(i) 8x³ – 6x² + 10x
(ii) 14mn + 22m – 62p
Solution:
(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
Solution:
(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)
Solution:
(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)
Solution:
(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)
Solution:
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]

ML Aggarwal Class 8 Solutions for ICSE Maths