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Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 4

Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 5
Ex 2.4, 5 - Chapter 2 Class 9 Polynomials - Part 6

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Ex 2.4, 5 Factorise: (ii) x3 – 3x2 − 9x − 5 Let p(x) = x3 – 3x2 – 9x – 5 Checking p(x) = 0 So, at x = –1, p(x) = 0 Hence, x + 1 is a factor of p(x) Now, p(x) = (x + 1) g(x) ⇒ g(x) = (𝑝(𝑥))/((𝑥+ 1)) ∴ g(x) is obtained after dividing p(x) by x + 1 So, g(x) = x2 – 4x – 5 So, p(x) = (x + 1) g(x) = (x + 1) (x2 – 4x – 5) We factorize g(x) i.e. x2 – 4x – 5 x2 – 4x – 5 We factorize using the splitting the middle term method = x2 – 5x + x – 5 = x(x – 5) + 1 (x – 5) = (x + 1) (x – 5) So, p(x) = (x + 1)(x + 1)(x – 5)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.