ML Aggarwal Solutions Class 9 Math 4th Chapter Factorisation Exercise 4.3
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fourth Chapter Factorisation Exercise 4.3. APC Solution Class 9 Exercise 4.3.
(1) (i) 4x2 – 25y2
Solution:
4x2 – 25y2
= (2x)2 – (5y)2
= (2x + 5y) (2x – 5y)
(ii) 9x2 – 1
Solution:
9x2 – 1
= (3x)2 – 12
= (3x + 1) (3x – 1)
(2) (i) 150 – 6a2
Solution:
150 – 6a2
= 6 (25 – a2)
= 6 [(5)2 – a2]
= 6 (5+a) (5-a)
(ii) 32x2 – 18y2
Solution:
32x2 – 18y2
= (16x2 – 9y2)
= 2 [(4x)2 – (3y)2]
= 2 (4x + 3y) (4x – 3y)
(3) (i) (x – y)2 – 9
Solution:
(x – y)2 – 9
= (x – y)2 – 32
= (x – y + 3) (x – y – 3)
(ii) 9 (x + y)2 – x2
Solution:
9 (x + y)2 – x2
= [3 (x + y)]2 – x2
= [3 (x + y) – x] [3 (x + y) + x]
= (3x + 3y – x) (3x + 3y + x)
= (2x + 3y) (4x + 3y)
(4) (i) 20x2 – 45y2
Solution:
20x2 – 45y2
= 5 (4x2 – 9y2)
= 5 [(2x)2 – (3y)2]
= 5 (2x + 3y) (2x – 3y)
(ii) 9x2 – 4 (y + 2x)2
Solution:
9x2 – 4 (y – 2x)2
= (3x)2 – [2 (y – 2x)]2
= (3x + 2y – 4x)
(3x – 2y + 4x)
= (2y – x) (7y – 2y)
(5) (i) 2 (x – 2y)2 – 50y2
Solution:
2 (x – 2y)2 – 50y2
= 2 [(x – 2y)2 – 25y2]
= 2 [(x – 2y)2 – (5y)2]
= 2 (x – 2y + 3y) (x – 2y – 5y)
= 2 (x + 3y) (x – 7y)
(ii) 32 – 2 (x – 4)2
Solution:
32 – 2 (x – 4)2
= 2 [16 – (x – 4)2]
= 2 [42 – (x – 4)2]
= 2 (4 + x – 4) (4 – x + 4)
= 2x (8 – x)
(6) (i) 108a2 – 3 (b – c)2
Solution:
108a2 – 3 (b – c)2
= 3 [36a2 – (b – c)2]
= 3 [(6a)2 – (b – c)2]
= 3 (6a – b + c) (6a + b + c)
(ii) πa5 – π3ab2
Solution:
πa5 – π3ab2
= πa (a4 – π2b2)
= πa [(a2)2 – (πb)2]
= πa (a2 – πb) (a2 + πb)
(7) (i) 50x2 – 2 (x – 2)2
Solution:
50x2 – 2 (x – 2)2
= 2 [25x2 – (x – 2)2]
= 2 [(5x)2 – (x – 2)2]
= 2 (5x + x – 2) (5x – x + 2)
= 2 (6x – 2) (4x + 2)
= 2 × 2 × 2 (3x – 1) (2x + 1)
= 8 (3x – 1) (2x + 1)
(ii) (x – 2) (x + 2) + 3
Solution:
(x – 2) (x + 2) + 3
= x2 – 4 + 3
= x2 – 1
= x2 – 12
= (x + 1) (x – 1)
(8) (i) x – 2y – x2 + 4y2
Solution:
x – 2y – x2 + 4y2
= x – 2y – 1 (x2 – 4y2)
= x – 2y – 1 [x2 – (2y)2]
= x – 2y – 1 (x – 2y) (x + 2y)
= (x – 2y) (1 – x – 2y)
(ii) 4a2 – b2 + 2a + b
Solution:
4a2 – b2 + 2a + b
= (2a)2 – b2 + 2a + b
= (2a + b) (2a – b) + 2a + b
= (2a + b) (2a – b + 1)
(9) (i) a (a – 2) – b (b – 2)
Solution:
a (a – 2) – b (b – 2)
= a2 – 2a – b2 – 2b
= a2 – b2 – 2a – 2b
= (a + b) (a – b) – 2 (a + b)
= (a + b) (a – b – 2)
(ii) a (a – 1) – b (b – 1)
Solution:
a (a – 1) – b (b – 1)
= a2 – a – b2 – b
= a2 – b2 – a – b
= (a + b) (a – b) – 1 (a + b)
= (a + b) (a – b – 1)
(10) (i) 9 – x2 + 2xy – y2
Solution:
9 – x2 + 2xy – y2
= 9 – 1 (x2 – 2xy + y2)
= 9 – 1 (x – y)2
= (3) – (x – y)2
= (3 + x – y) (3 – x + y)
(ii) 9x4 – (x2 + 2x + 1)
Solution:
= (3x2)2 – (x + 1)2
= (3x2 + x + 1) (3x2 – x – 1)
(11) (i) 9x4 – x2 – 12x – 36
Solution:
9x4 – x2 – 12x – 36
= 9x4 – 1 (x2 + 12x + 36)
= 9x4 – 1 (x2 + 2 × 6 × x + 62)
= 9x4 – 1 (x + 6)2
= 9x4 – (x + 6)2
= (3x2)2 – (x + 6)2
= (3x2 + x + 6) (3x2 – x – 6)
(ii) x3 – 5x2 – x + 5
Solution:
x3 – 5x2 – x + 5
= x2 (x – 5) – 1 (x – 5)
= (x – 5) (x2 – 1)
= (x – 5) (x + 1) (x – 1)
(12) a4– b4 + 2b2 – 1
Solution:
a4 – b4 + 2b2 – 1
= a4 – 1 (b4 – 2b2 + 1)
= a4 – 1 [(b2)2 – 2 × b2 × 1 + 12]
= a4 – 1 (b2 – 1)2
= a4 – (b2 – 1)2
= (a2)2 – (b2 – 1)2
= (a2 – b2 + 1) (a2 + b2 – 1)
= [(a + b) (a – b) + 1] [a2 + (b + 1) (b – 1)]
(ii) x3 – 25x
Solution:
x3 – 25x
= x (x2 – 25)
= x [x2 – 52]
= x (x + 5) (x – 5)
(13) (i) 2x4 – 32
Solution:
2x4 – 32
= 2 (x4 – 16)
= 2 [(x2)2 – 42]
= 2 (x2 + 4) (x2 – 4)
= 2 (x2 + 4) [x – 22]
= 2 (x2 + 4) (x + 2) (x – 2)
(ii) a2 (b + c) – (b + c)3
Solution:
a2 (b + c) – (b + c)3
= (b + c) [a2 – (b + c)2]
= (b + c) (a + b + c) (a – b – c)
(14) (i) (a + b)3 – a – b
Solution:
(a + b)3 – a – b
= (a + b)3 – 1 (a + b)
= (a + b) [(a + b)2 – 1]
= (a + b) (a + b + 1) (a + b – 1)
(ii) x2 – 2xy + y2 – a2 – 2ab – b2
Solution:
x2 – 2xy + y2 – a2 – 2ab – b2
= (x – y)2 – 1 (a2 + 2ab + b)2
= (x – y)2 – (a + b)2
= (x – y + a + b) (x – y – a – b)
(15) (i) (a2 – b2) (c2 – d2) – 4abcd
Solution:
(a2 – b2) (c2 – d2) – 4abcd
= a2c2 – a2d2 – b2c2 + b2d2 – 4abcd
= a2c2 – 4abcd + b3d3 – a2d3 – b2c2
= a2c2 – 2abcd + b2 d2 – a3d2 – 2abcd – b3c2
= a3c3 – 2abcd + b2d2 – 1 (a2d2 + 2abcd + b2c2)
= (ac)2 – 2ac.bd + (bd)2 -1 [(ad)2 + 2ad.bc + (bc)2]
= (ac – bd)2– (ad + bc)2
= (ac – bd + ad + bc) (ac – bd – ad – bc)
(15) (ii) 4x2 – y2 – 3xy + 2x – 2y
Solution:
4x2 – y2 – 3xy + 2x – 2y
= x2 – y2 + 3x2 – 3xy + 2x – 2y
= (x + y) (x – y) + 3x (x – y) + 2 (x – y)
= (x – y) (x + y + 3x + 2)
= (x – y) (4x + y + 2)
(16) (i) x2 + 1/x2 – 11
Solution:
x2 + 1/x2 – 11
= x2 + 1/x2 – 2 – 9
= (x – 1/x)2– 9
= (x – 1/x)2 – 32
= (x – 1/x + 3) (x – 1/x – 3)
(ii) x4 + 5x2 + 9
Solution:
x4 + 5x2 – 9
= x4 + 6x2 + 9 – x2
= (x2)2 + 2.x2× 3 + 32 – x2
= (x2 + 3)2 – x2
= (x2 + 3 + x) (x2 + 3 – x)
(17) (i) a4 + b4 – 7a2b2
Solution:
a4 + 2a2b2 + b2 – 9a2b2
= (a2 + b2)2 – (3ab)2
= (a2 + b2 + 3ab) (a2 + b2 – 3ab) – Ans.
= (a2 + b2 + 2ab + ab) (a2 + b2 – 2ab – ab)
= [(a + b)2 + ab] [(a – b)2 – ab]
= ab [(a + b)2 + 1] [(a – b)2 – 1]
= ab [(a + b)2 + 1] [(a + b) (a – b) – 1]
(ii) x4 – 14x2 + 1
Solution:
x4– 14x2 + 1
= x4 + 2x2 + 1 – 16x2
= (x2)2 + 2.x2 + 12 – (4x)2
= (x2 + 1)2 – (4x)2
= (x2 + 1 – 4x) (x2 + 1 + 4x)
(18) Express each of the following as the difference of two squares:
(i) (x2 – 5x + 7) (x2 + 5x + 7)
Solution:
(x2 – 5x + 7) (x2 + 5x + 7)
= (x2 + 7 – 5x) (x2 + 7 + 5x)
= (x2 + 7)2 – (5x)2
(ii) (x2 – 5x + 7) (x2 – 5x – 7)
Solution:
(x2 – 5x + 7) (x2 – 5x – 7)
= (x2 – 5x)2 – 72
(iii) (x2+ 5x – 7) (x2 – 5x + 7)
Solution:
(x2+ 5x – 7) (x2 – 5x + 7)
= [(x2 + 1 (5x – 7)] [x2 – 1) (5x – 7)]
= [x2 + (5x – 7)] [x2 – (5x – 7)]
= (x2)2 – (5x – 7)2
(19) Evaluate the following by using factors:
(i) (979)2 – (21)2
Solution:
(979)2 – (21)2
= (979 + 21) (979 – 21)
= 1000 × 958
= 958000
(ii) (99.9)2 – (0.1)2
Solution:
(99.9)2 – (0.1)2
= (99.9 + 1) (99.9 – 0.1)
= 100 × 99.8
= 9980
Yes,very best answers.