Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 10

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Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 11

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Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 12

  1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise

Transcript

Ex 2.4, 5 Factorise: (iv) 2y3 + y2 − 2y − 1 Let p(y) = 2y3 + y2 – 2y – 1 Checking p(y) = 0 So, at y = 1, p(y) = 0 Hence, y – 1 is a factor of p(y) Now, p(y) = (y – 1) g(y) ⇒ g(y) = (𝑝(𝑦))/((𝑦 − 1)) ∴ g(y) is obtained after dividing p(y) by y – 1 So, g(y) = 2y2 + 3y + 1 So, p(y) = (y – 1) g(y) = (y – 1) (2y2 + 3y + 1) We factorize g(y) i.e. 2y2 + 3y + 1 2y2 + 3y + 1 We factorize using the splitting the middle term method = 2y2 + 2y + y + 1 = 2y(y + 1) + 1 (y + 1) = (y + 1) (2y + 1) So, p(y) = (y – 1)(y + 1)(2y + 1)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.