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

**Factorise the following (1 to 8) polynomials:**

Question 1.

(i) 8xy^{3} + 12x^{2}y^{2}

(ii) 15ax^{3} – 9ax^{2}

Solution:

(i) 8xy^{3} + 12x^{2}y^{2} = 4xy^{2} (2y + 3x)

(ii) 15ax^{3} – 9ax^{2} = 3ax^{2} (5x – 3)

Question 2.

(i) 21 py^{2} – 56py

(ii) 4x^{3} – 6x^{2}

Solution:

(i) 21 py^{2} – 56py = 7py (3y – 8)

(ii) 4x^{3} – 6x^{2} = 2x^{2} (2x – 3)

Question 3.

(i) 25abc^{2} – 15a^{2}b^{2}c

(ii) x^{2}yz + xy^{2}z + xyz^{2}

Solution:

(i) 25abc^{2} – 15a^{2}b^{2}c = 5abc (5c – 3ab)

(ii) x^{2}yz + xy^{2}z + xyz^{2} = xyz(x + y + z)

Question 4.

(i) 8x^{3} – 6x^{2} + 10x

(ii) 14mn + 22m – 62p

Solution:

(i) 8x^{3} – 6x^{2} + 10x = 2x (4x^{2} – 3x + 5)

(ii) 14mn + 22m – 62p = 2 (7mn + 11m – 31p)

Question 5.

(i) 18p^{2}q^{2} – 24pq^{2} + 30p^{2}q

(ii) 27a^{3}b^{3} – 18a^{2}b^{3} + 75a^{3}b^{2}

Solution:

(i) 18p^{2}q^{2} – 24 pq^{2} + 30p^{2}q

= 6pq (3pq -4q + 5p)

(ii) 27a^{3}b^{3} – 18a^{2}b^{3} + 75a^{3}b^{2}

= 3a^{2}b^{2} (9ab – 6b + 25a)

Question 6.

(i) 15a (2p – 3p) – 106 (2p – 3q)

(ii) 3a (x^{2} + y^{2}) + 6b (x^{2} + y^{2})

Solution:

(i) 15a (2p – 3q) – 10b (2p – 3q)

= (2p – 3q)(15a – 10b)

= (2p – 3q) (5) (3a – 2b)

= 5 (2p- 3q) (3a – 2b)

(ii) 3a (x^{2} + y^{2}) + 66 (x^{2} + y^{2})

= (x^{2} + y^{2}) (3a + 6b)

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

= 3 (x^{2} + y^{2}) (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]