Ex 2.5, 3 - Factorise following using appropriate identities - Ex 2.5

  1. Chapter 2 Class 9 Polynomials
  2. Serial order wise
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Ex 2.5, 2 Evaluate the following products without multiplying directly: (iii) 104 × 96 104 × 96 = (100 + 4) (100 − 4) Using the identity  (x + y) (x – y) = x2 – y2 where x = 100 , y = 4 = (100)2 − (4)2  = 10000 − 16 = 9984 Ex 2.5, 3 Factorise the following using appropriate identities: (i) 9x2 + 6xy + y2 9x2 + 6xy + y2 = 32 x2 + 6xy +(y)2 = (3x)2 + 6xy +(y)2 = (3x)2 + 2 (3x) (y) +(y)2 Using Identity (a + b)2 = a2 + b2 + 2ab Where a = 3x , b = y = (3x + y)2 = (3x + y) (3x + y) Ex 2.5, 3 Factorise the following using appropriate identities: (ii) 4y2 – 4y + 1 4y2 – 4y + 1 = 22 y2 – 4y + 12 = (2y)2 – 4y + 12 = (2y)2 – 2 (2y) (1) + (1)2 Using Identity (a – b)2 = a2 + b2 – 2ab Where a = 2y , b = 1 = (2y – 1)2 = (2y – 1) (2y – 1) Ex 2.5, 3 Factorise the following using appropriate identities: (iii) x2 – 𝑦2/100 x2 – 𝑦2/100 = x2 – 𝑦2/〖10〗^2 = x2 – (y/10)^2 Using identity (a + b) (a – b) = a2 – b2 Where a = x , b = 𝑦/10 = (x+ y/10) (x− y/10)

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