ML Aggarwal Factorisation Exe-4.1 Class 9 ICSE Maths Solutions

ML Aggarwal Factorisation Exe-4.1 Class 9 ICSE Maths Solutions . Step by step solutions  of  Factorisation problems as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Factorisation Exe-4.1 Class 9 ICSE Maths Solutions

Board	ICSE
Subject	Maths
Class	9th
Chapter-4	Factorisation
Topics	Solution of Exe-4.1 Questions
Academic Session	2024-2025

Solution of Exe-4.1 Questions

ML Aggarwal Factorisation Exe-4.1 Class 9 ICSE Maths Solutions

Factories the following :

Que-1:

(i) 8xy³ + 12x²y²

(ii) 15 ax³ – 9ax²

Sol: (i) 8xy³ + 12x²y²

Take out common in both terms,

Then, 4xy² (2y + 3x)

Therefore, HCF of 8xy³ and 12x²y² is 4xy².

(ii) 15 ax³ – 9ax²

Take out common in both terms,

Then, 3ax² (5x – 3)

Therefore, HCF of 15 ax³ and 9ax² is 3ax².

Que-2:

(i) 21py² – 56py

(ii) 4x³ – 6x²

Sol:  (i) 21py² – 56py

Take out common in both terms,

Then, 7py (3y – 8)

Therefore, HCF of 21py² and 56py is 7py.

(ii) 4x³ – 6x²

4x³ – 6x²

Take out common in both terms,

Then, 2x² (2x – 3)

Therefore, HCF of 4x³ and 6x² is 2x².

Que-3:

(i) 2πr² – 4πr

(ii) 18m + 16n

Sol: (i) 2πr² – 4πr

Take out common in both terms,

Then, 2πr (r – 2)

Therefore, HCF of 2πr² and 4πr is 2πr.

(ii) 18m + 16n

18m + 16n

Take out common in both terms,

Then, 2 (9m + 8n)

Therefore, HCF of 18m and 16n is 2.

Que-4:

(i) 25abc² – 15a²b²c

(ii) 28p²q²r – 42pq²r²

Sol: (i) 25abc² – 15a²b²c

Take out common in both terms,

Then, 5abc (5c – 3ab)

Therefore, HCF of 25abc² and 15a²b²c is 5abc.

(ii) 28p²q²r – 42pq²r²

28p²q²r – 42pq²r²

Take out common in both terms,

Then, 14pq²r (2p – 3r)

Therefore, HCF of 28p²q²r and 42pq²r² is 14pq²r.

Que-5:

(i) 8x³ – 6x² + 10x

(ii) 14mn + 22m – 62p

Sol: (i) 8x³ – 6x² + 10x

Take out common in all terms,

Then, 2x(4x² – 3x + 5)

Therefore, HCF of 8x³, 6x² and 10x is 2x.

(ii) 14mn + 22m – 62p

14mn + 22m – 62p

Take out common in all terms,

Then, 2 (7mn + 11m – 31p)

Therefore, HCF of 14mn, 22m and 62p is 2.

Que-6:

(i) 18p²q² – 24pq² + 30p²q

(ii) 27a³b³ – 18a²b³ + 75a³b²

Sol: (i) 18p²q² – 24pq² + 30p²q

Take out common in all terms,

Then, 6pq (3pq – 4q + 5p)

Therefore, HCF of 18p²q², 24pq² and 30p²q is 6pq.

(ii) 27a³b³ – 18a²b³ + 75a³b²

27a³b³ – 18a²b³ + 75a³b²

Take out common in all terms,

Then, 3a²b² (9ab – 6b + 25a)

Que-7:

(i) 15a (2p – 3q) – 10b (2p – 3q)

(ii) 3a(x² + y²) + 6b (x² + y²)

Sol: (i) 15a (2p – 3q) – 10b (2p – 3q)

Take out common in both terms,

Then, 5(2p – 3q) [3a – 2b]

Therefore, HCF of 15a (2p – 3q) and 10b (2p – 3q) is 5(2p – 3q).

(ii) 3a(x² + y²) + 6b (x² + y²)

3a(x² + y²) + 6b (x² + y²)

Take out common in all terms,

Then, 3(x² + y²) (a + 2b)

Therefore, HCF of 3a(x² + y²) and 6b (x² + y²) is 3(x² + y²).

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

Sol:   (i) 6(x + 2y)³ + 8(x + 2y)²

Take out common in all terms,

Then, 2(x + 2y)² [3(x + 2y) + 4]

Therefore, HCF of 6(x + 2y)³ and 8(x + 2y)² is 2(x + 2y)².

(ii) 14(a – 3b)³ – 21p(a – 3b)

14(a – 3b)³ – 21p(a – 3b)

Take out common in all terms,

Then, 7(a – 3b) [2(a – 3b)² – 3p]

Therefore, HCF of 14(a – 3b)³ and 21p(a – 3b) is 7(a – 3b).

Que-9: (i) 10a(2p + q)³ – 15b (2p + q)² + 35 (2p + q), (ii) x(x² + y² – z²) + y(-x² – y² + z²) – z (x² + y – z²)

Sol:  (i) 10a(2p + q)³ – 15b (2p + q)² + 35 (2p + q)

Take out common in all terms,

Then, 5(2p + q) [2a (2p + q)² – 3b (2p + q) + 7]

Therefore, HCF of 10a(2p + q)³, 15b (2p + q)² and 35 (2p + q) is 5(2p + q).

(ii) x(x² + y² – z²) + y(-x² – y² + z²) – z (x² + y – z²)

x(x² + y² – z²) + y(-x² – y² + z²) – z (x² + y – z²)

Take out common in all terms,

Then, (x² + y² – z²) [x – y – z]

Therefore, HCF of x(x² + y² – z²), y(-x² – y² + z²) and z (x² + y – z²) is (x² + y² – z²)

—  : End of ML Aggarwal Factorisation Exe-4.1 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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