ML Aggarwal Factorisation Exe-11.3 Class 8 ICSE Ch-11 Maths Solutions. We Provide Step by Step Answer of Exe-11.3 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-11.3 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-11.3 Questions |
| Edition | 2023-2024 |
Factorisation Exe-11.3
ML Aggarwal Class 8 ICSE Maths Solutions
Page-196
Question 1. Factories the following expressions using algebraic identities:
(i) x2 – 12x + 36
(ii) 36p2 – 60pq + 25q2
(iii) 9y2 + 66xy + 121y2
(iv) a4 + 6a2b2 + 9b4
(v) x2 + 1/x2 + 2
(vi) x2 + x + 1/4
Answer:
Using (a + b)2 = a2 + 2ab +b2 and (a – b)2 = a2 – 2ab + b2
(i) y2 – 12x + 36
= (x)2 – 2 × x × 6 + (6)22
= (x – 6)2
(ii) 36p2 – 60pq + 25q2
= (6p)2 – 2 × 6p × 5q + (5q)2
= (6p – 5q)2
(iii) 9x2 + 66xy + 121 y2
= (3x)2 + 2 × 3x × 11y + (11y)2
= (3x + 11 y)2
(iv) a4 + 6a2b2 + 9b4
= (a2)2 + 2 × 2a2 × 3b2 + (3b2)2
= (a2 + 3b2)2
(v) x2 + 1/x2 + 2
= (x)2 + 2 × x × 1/x + (1/x)2
= (x + 1/x)2
(vi) x2 + x + ¼
= (x)2 + 2 × x × 1/2 + (1/2)2
= (x + 1/2)2
Factorisation Exe-11.3
ML Aggarwal Class 8 ICSE Maths Solutions
Page-197
Factorise the following (2 to 13) expressions:
Question 2.
(i) 4p2 – 9
(ii) 4x2 – 169y2
Answer:
(i) 4p2 – 9
= (2p)2 – (3)2
= (2p + 3) (2p – 3)
(ii) 4x2 – 169y2
= (2x)2 – (13y)2
= (2x + 13y) (2x – 13y)
Question 3.
(i) 9x2y2 – 25
(ii) 16×2 – 1/144
Answer:
(i) 9x2y2 – 25
= (3xy)2 – (5)2
= (3xy + 5) (3xy – 5)
(ii) 16x2 – 1/144
= (4x)2 – (1/12)2
= (4x + 1/12) (4x – 1/12)
Question 4.
(i) 20×2 – 45y2
(ii) 9/16 – 25a2b2
Answer:
(i) 20x2 – 45y2
= 5 (4x2 – 9y2)
= 5[(2x)2 – (3y)2]
= 5 (2x + 3y) (2x – 3y)
(ii) 9/16 – 25a2b2
= (¾)2 – (5ab)2
= (¾ + 5ab) (¾ – 5ab)
Question 5.
(i) (2a + 3b)2 – 16c2
(ii) 1 – (b – c)2
Answer:
(i) (2a + 3b)2 – 16c2
= (2a + 3b)2 – (4c)2
= (2a + 3b + 4c) (2a + 3b – 4c)
(ii) 1 – (b – c)2
= (1)2 – (b – c)2
= [1 + b – c)] [1 – (b – c)]
= (1 +b – c)(1 – b + c)
Question 6.
(i) 9 (x + y)2 – x2
(ii) (2m + 3n)2 – (3m + 2n)2
Answer:
(i) 9 (x + x)2 – x2
= [3 (x + y)]2 – [x]2
= [3 (x + y) + x] [3 (x + y) – x]
= (3x + 3y + x) (3x + 3y – x)
= (4x + 3y) (2x + 3x)
(ii) (2m + 3n)2 – (3m + 2n)2
= (4m2 + 9n2 + 12mn) – (9m2 + 4n2 + 12mn)
= 4m2 + 9n2 + 12mn – 9m2 – 4m2 – 12mn
= 4m2 + 9n2 – 9m2 – 4n2
= – 5m2 + 5n2
= 5 (n2 – m2)
= 5 (m + n) (n – m)
Question 7.
(i) 25 (a + b)2 – 16 (a – b)2
(ii) 9 (3x + 2)2 – 4 (2x – 1)2
Answer:
(i) 25 (a + b)2 – 16 (a – b)2
= [5 (a + b)]2 – [4 (a – b)]2
= (5a + 5b)2 – (4a – 4b)2
= [(5a + 5b)2 + (4a – 4b)] [(5a + 5b) – (4a – 4b)]
= (5a + 5b + 4a – 4b) (5a + 5b – 4a + 4b)
= (9a + b) (a + 9b)
(ii) 9 (3x + 2)2 – 4 (2x – 1)2
= [3 (3x + 2)]2 – [2 (2x – 1)]2
= (9x + 6)2 – (4x – 2)2
= [(9x + 6) + (4x – 2)] [(9x + 6) – (4x – 2)]
= (9x + 6 + 4x – 2) (9x + 6 – 4x + 2)
= (13x + 4) (5x + 8)
Question 8.
(i) x3 – 25x
(ii) 63p2q2 – 7
Answer:
(i) x3 – 25x
= x (x2 – 25) = x [(x)2 – (5)2]
= x (x + 5) (x – 5)
(ii) 63p2q2 – 7
= 7 (9p2q2 – 1)
= 7 [(3pq)2 – (1)2]
= 7 (3pq + 1) (3pq – 1)
Question 9.
(i) 32a2b – 72b3
(ii) 9 (a + b)3 – 25 (a + b)
Answer:
(i) 32 a2b – 72b3
= 8b (4a2 – 9b2) ⇒ 8b [(2a)2 – (3b)2]
= 8b (2a + 3b) (2a – 3b)
(ii) 9 (a + b)3 – 25 (a + b)
= (a + b) [9 (a + b)2 – 25]
= (a + b) [{3 (a + b)}2 – (5)2]
= (a + 6) [(3a + 3b)2 – (5)2]
= (a + b) [(3a + 3b + 5) (3a + 36 – 5)]
= (a + b) (3a + 3b + 5) (3a + 3b – 5)
Question 10.
(i) x2 – y2 – 2y – 1
(ii) p2– 4pq + 4q2 – r2
Answer:
(i) x2 – y2 – 2y – 1
= x2 – (y2 + 2y + 1)
= (x)2 – (y + 1)2
= [x + (y + 1)] [x – (y + 1)]
= (x + y + 1) (x – y – 1)
(ii) p2 – 4pq + 4q2 – r2
= (p)2 – 2 × p × 2q + (2q)2 – r2 [∵ (a – b)2 = a2 – 2ab + b2]
= (p – 2q)2 – (r)2
= (p – 2q + r)(p – 2q – r) [∵ a2 – b2 = (a + b)(a – b)]
Question 11.
(i) 9x2 – y2 + 4y – 4
(ii) 4a2 – 4b2 + 4a + 1
Answer:
(i) 9x2 – y2 + 4y – 4
= 9x2 – (y2 – 4y + 4)
= 9x2 – (y – 2)2
= (3x)2 (y – 2)2
= [3x + (y – 2)] [3x – (y – 2)]
= (3x + y – 2) (3x – y + 2)
(ii) 4a2 – 4b2 + 4a + 1
= (4a2 + 4a + 1) – 4b2
= (2a + 1)2 – (2b)2
= (2a + 2b + 1) (2a – 2b + 1)
Question 12.
(i) 625 – p4
(ii) 5y5 – 405y
Answer:
(i) 625 – p4
= (25)2 – (p2)2
= (25 + p2) (25 – p2)
= (25 + p2) [(5)2 – (p)2]
= (25 +p2) (5 + p) (5 – p)
(ii) 5y5 – 405y
= 5y(y4 – 81)
= 5y [(y2)2 – (9)2]
= 5y (y2 + 9) (y2 – 9)
= 5y (y2 + 9) [(y)2 – (3)2
= 5y (y2 + 9) (y + 3) (y – 3)
(ML Aggarwal Factorisation Exe-11.3 Class 8 ICSE Maths)
Question 13.
(i) x4 – y4 + x2 – y2
(ii) 64a2 – 9b2 + 42bc – 49c2
Answer:
(i) x4 – y4 + x2 – y2
= [(x2)2 – (y2)2] + (x2 – y2) [Using, a2 – b2 = (a + b) (a – b)]
= (x2 + y2) (x2 – y2) + 1(x2 – y2)
= (x2 – y2) (x2 + y2 + 1)
= (x + y(x – y)(x2 + y2 + 1)
(ii) 64a2 – 9b2 + 42bc – 49c2
= 64a2 – [9b2 – 42bc + 49c2]
= (8a)2 – [(3b)2 – 2 × 3b × 7c + (7c)2] [∵ a2 + b2 – 2ab = (a – b)2 and a2 – b2 = (a + b)(a – b)]
= (8a)2 – (3b – 7c)2
= (8a + 3b – 7c) (8a – 3b + 7c)
Question 14.
Evaluate the following:
(i) {(38)2 – (22)2}/16
(ii) (5.3)2 – (3.3)2/{(11.2)2 – (11.2) x (2.4) + (1.2)2 }
Answer:
(i) Since a2−b2 = (a+b)(a−b), therefore
382−222
=(38−22)(38+22)
=16×60
So, (382−222)/16
= (16×60)/16
= 60
(ii) (5.3)2 – (3.3)2/{(11.2)2 – (11.2) x (2.4) + (1.2)2 }
Since a2 – b2 = (a+b)(a-b)
= (5.3+3.3)(5.3-3.3) / (11.2)2 -(11.2)×(2.4)+ (1.2)2
= (8.6×2) / (11.2)2– 11.2× (2.4) + (1.2)2
Since (a-b)2 = a2 -2ab + b2
= 17.2 / (11.2)2 – 2(11.2 ×1.2) + (1.2)2
=17.2 / (11.2 – 1.2)2
=17.2 / (10)2
=17.2 / 100
=172/1000
=0.172
— End of Factorisation Exe-11.3 Class 8 ICSE Maths Solutions :–
Return to : – ML Aggarwal Maths Solutions for ICSE Class -8
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