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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 2.3, 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.