ML Aggarwal Solutions Class 9 Math 4th Chapter Factorisation Exercise 4.1
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fourth Chapter Factorisation Exercise 4.1. APC Solution Class 9 Exercise 4.1.
(1) (i) 8xy3 + 12x2y2
Solution:
8xy3 + 12x2y2
HCF of 8xy3 & 12x2y2 is 4xy2
∴ 8xy3 + 12x2y2 = 4xy2 (2y + 3x)
(ii) 15ax3 – 9ax2
Solution:
HCF of 15ax3 & 9ax2 is 3ax2
∴ 15ax3 – 9ax2 = 3ax2 (5x – 3)
(2) (i) 21py2 – 56py
Solution:
21py3 – 56py
HCF of 21py2 & 56py = 7py
∴ 21py2 – 56py = 7py (3y – 8)
(ii) 4x3 – 6x2
Solution:
4x3 – 6x2
HCF of 4x3 & 6x2 is 2x2
∴ 4x3 – 6x2 = 2x2 (2x – 3)
(3) (i) 2πr2 – 4πr
Solution:
2πr2 – 4πr
HCF of 2πr2 & 4πr is 2πr
∴ 2πr2 – 4πr = 2πr (r – 2)
(ii) 18m + 16n
Solution:
18m + 16n
HCF of 18m & 16n is 2
∴ 18m + 16n = 2 (9m + 8n)
(4) (i) 25abc2 – 15a2b2c
(ii) 28p2q2r – 42pq2r2
Solution:
(i) 25abc2 – 15a2b2c
HCF of 25abc2 & 15a2b2c is 5abc
∴ 25abc2 – 15a2b2c = 5abc (5c – 3ab)
(ii) 28p2q2r – 42pq2r2
HCF of 28p2q2& 42pq2r2 is 14pq2r
∴ 28p2q2r – 42pq2r2 = 14pq2r (2p – 3r)
(5) (i) 8x3 – 6x2 + 10x
(ii) 14mn + 22m – 62p
Solution:
(i) 8x3 – 6x2 + 10x
HCF of 8x3, 6x2& 10x is 2x
∴ 8x3 – 6x3 + 10x = 2 × (4x2– 3x + 5)
(ii) 14mn + 22m – 62p
HCF 14mn, 22m, 62p is 2
∴ 14mn + 22m – 62p = 2 (7mn + 11m – 31p)
(6) (i) 18p2q2 – 24pq2 + 30p2q
(ii) 27a3b3 – 18a2b3 + 75a3b2
Solution:
(i) 18p2q2 – 24pq2 + 30p2q
HCF of 18p2q2, 24pq2 & 30p2q is 619
∴ 18p2q2 – 24pq2 + 30p2q = 6pq (3pq – 49 + 5p)
(ii) 27a3b3 – 18a3b3 + 75a3b2
HCF of 27a3b3, 18a2b3 & 75a3b2 is 3a2b2
∴ 27a3b3 – 18a2b3 + 75a3b2 = 3a2b2 (9ab – 66 + 25a)
(7) (i) 15a (2p – 3q) – 10b (2p – 3q)
(ii) 3a (x2 + y2) + 6b (x2 + y2).
Solution:
(i) 15 (2p – 3q) – 10b (2p – 3q)
HCF of 15 (2p – 3q) & 10b (2p – 3q) is 5 (2p – 3q)
∴ 15 (2p – 3q) – 10b (2p – 3q) = 5 (2p – 3q) (3 – 2b)
(ii) 3a (x2 + y2) + 6b (x2 + y2)
HCF of 3a (x2 + y2) & 6b (x2 + y2) = 3 (x2 + y2) (a + 2b)
(8) (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
HCF of 6 (x + 2y)3, 8 (x + 2y)2 is 2 (x + 2y)2
∴ 6 (x + 2y)3+ 8 (x + 2y)2 = 2 (x + 2y)n [3 (x + 2y) + 4]
(ii) 14 (a – 3b)3 – 21 p (a – 3b)
HCF of 14 (a – 3b)3& 21p (a – 3b) is 7 (a – 3b)
∴ 14 (a – 3b)3 – 21 p (a – 3b) = 7 (a – 3b) [2 (a – 3b)2 – 3b]
(9) (i) 10a (2p + q)3 – 15b (2p + q)2 + 35 (2p + q)
Solution:
10a (2p + q)3 – 15b (2p + q)2 + 35 (2p + q)
HCF of 10a (2p + q)3, 15b (2p + q)2& 35 (2p + q) is 5 (2p + q)
∴ 10a (2p + q)3 – 15b (2p + q)2 + 35 (2p + q) = 5 (2p + q) [2a (2p + q)2 – 3b (2p + q) + 7]
(ii) x (x2 – y2 – z2) + y (-x2 – y2 + z2) – z (x2 + y2 – z2)
Solution:
x (x2 – y2 – z2 ) + y (-x2 – y2 + z2) – z (x2 + y2 – z2)
= x (x2 + y2 – z2) – y (x2 + y2 – z2) – z (x2 + y2 – z2)
HCF of x (x2 + y2 – z2), y (x2 + y2 – z2), z (x2 + y2 – z2) is (x2 + y2 – z2)
∴ x (x2 + y2 – z2) – y (x2 + y2 – z2) – z (x2 + y2 – z2)
∴ x (x2 + y2 – z2) – y (x2 + y2 – z2) – z (x2 + y2 + z2)
= (x2 + y2 – z2) (x – y – z)