RS Aggarwal Class 8 Math Seventh Chapter Factorisation Exercise 7B Solution

RS Aggarwal Class 8 Math Seventh Chapter Factorisation Exercise 7B Solution

EXERCISE 7B

Factorise:

FORMULA: (a2 – b2) = (a + b) (a – b)

(1) x2 – 36

= x2 – 62

= (x – 6) (x + 6)

(2) 4a2 – 9

= (2a)2 – (3)2

= (2a – 3) (2a + 3)

(3) 81 – 49x2

= (9)2 – (7x)2

= (9 – 7x) (9 + 7x)

(4) 4x2 – 9y2

= (2x)2 – (3y)2

= (2x – 3y) (2x + 3y)

(5) 16a2 – 225b2

= (4a)2 – (15b)2

= (4a – 15b) (4a + 15b)

(6) 9a2b2 – 25

= (3ab)2 – (5)2

= (3ab – 5) (3ab + 5)

(7) 16a2 – 144

= (4a)2 – (12)2

= (4a – 12) (4a + 12)

= [4(a – 3)] [4(a + 3)]

= (4 × 4) (a – 3) (a + 3)

= 16 (a – 3) (a + 3)

(8) 63a2 – 112b2

= 7(9a2 – 16b2)

= 7 [(3a)2 – (4b)2]

= 7 (3a – 4b) (3a + 4b)

(9) 20a2 – 45b2

= 5(4a2 – 9b2)

= 5 [(2a)2 – (3b)2]

= 5 (2a – 3b) (2a + 3b)

(10) 12x2 – 27

= 3(4x2 – 9)

= 3[(2x)2 – (3)2]

= 3 (2x – 3) (2x + 3)

(11) x3 – 64x

= x[(x)2 – (8)2]

= x (x – 8) (x + 8)

(12) 16x5 – 144x3

= 16x3 (x2 – 9)

= 16x3 [(x)2 – (3)2]

= 16x3 (x – 3) (x + 3)

(13) 3x5 – 48x3

= 3x3[(x)2 – 16]

= 3x3[(x)2 – (4)2]

= 3x3 (x – 4) (x + 4)

(14) 16p3 – 4p

= 4p (4p2 – 1)

= 4p [(2p)2 – (1)2]

= 4p (2p – 1) (2p + 1)

(15) 63a2b2 – 7

= 7 (9a2b2 – 1)

= 7 [(3ab)2 – (1)2]

= 7 (3ab – 1) (3ab + 1)

(16) 1 – (b – c)2

= (1)2 – (b – c)2

= (1 – b + c) (1 + b – c)

(17) (2a + 3b)2 – 16c2

= (2a + 3b)2 – (4c)2

= (2a + 3b – 4c) (2a + 3b + 4c)

(18) (l + m)2 – (l – m)2

= (l + m + l – m) (l + m – l + m)

= 2l × 2m = 4lm

(19) (2x + 5y)2 – 1

= (2x + 5y)2 – (1)2

= (2x + 5y – 1) (2x + 5y + 1)

(20) 36c2 – (5a + b)2

= (6c)2 – (5a + b)2

= (6c + 5a + b) (6c – 5a – b)

(21) (3x – 4y)2 – 25z2

= (3x – 4y)2 – (5z)2

= (3x – 4y + 5z) (3x – 4y – 5z)

(22) x2 – y2 – 2y – 1

= x2 – (y2 + 2y + 1)

= x2 – (y + 1)2

= (x + y + 1) (x – y – 1)

(23) 25 – a2 – b2 – 2ab

= 25 – (a2 + 2ab + b2)

= (5)2 – (a + b)2

= (5 + a + b) (5 – a – b)

(24) 25a2 – 4b2 + 28bc – 49c2

= 25a2 – [(2b)2 – (2 × 2b × 7c) + (7c)2]

= (5a)2 – (2b – 7c)2

= (5a + 2b – 7c) (5a – 2b + 7c)

(25) 9a2 – b2 + 4b – 4

= 9a2 – [(b)2 – (2 × b × 2) + (2)2]

= (3a)2 – (b – 2)2

= (3a + b – 2) (3a – b + 2)

(26) 100 – (x – 5)2

= (10)2 – (x – 5)2

= (10 + x – 5) (10 – x + 5)

= (5 + x) (15 – x)

(27) Evaluate {(405)2 – (395)2}

= (405 + 395) (405 – 395)

= 800 × 10 = 8000

(28) Evaluate {(7.8)2 – (2.2)2}

= (7.8 + 2.2) (7.8 – 2.2)

= 10 × 5.6 = 56


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