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Last updated at Aug. 25, 2021 by Teachoo

Transcript

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)

Ex 2.4

Ex 2.4, 1 (i)
Important

Ex 2.4, 1 (ii)

Ex 2.4, 1 (iii)

Ex 2.4, 1 (iv) Important

Ex 2.4, 2 (i)

Ex 2.4, 2 (ii) Important

Ex 2.4, 2 (iii)

Ex 2.4, 3 (i)

Ex 2.4, 3 (ii) Important

Ex 2.4, 3 (iii) Important

Ex 2.4, 3 (iv)

Ex 2.4, 4 (i)

Ex 2.4, 4 (ii) Important

Ex 2.4, 4 (iii)

Ex 2.4, 4 (iv) Important

Ex 2.4, 5 (i)

Ex 2.4, 5 (ii) Important You are here

Ex 2.4, 5 (iii) Important

Ex 2.4, 5 (iv)

About the Author

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.