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

Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 11
Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 12

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Transcript

Ex 2.3, 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.