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Ex 2.3, 5 (iv) - Chapter 2 Class 9 Polynomials
Last updated at May 29, 2023 by Teachoo
Ex 2.3
Ex 2.3, 1 (i)
Important
Deleted for CBSE Board 2024 Exams
Ex 2.3, 1 (ii)
Ex 2.3, 1 (iii)
Ex 2.3, 1 (iv) Important
Ex 2.3, 2 (i) Deleted for CBSE Board 2024 Exams
Ex 2.3, 2 (ii) Important
Ex 2.3, 2 (iii)
Ex 2.3, 3 (i) Deleted for CBSE Board 2024 Exams
Ex 2.3, 3 (ii) Important
Ex 2.3, 3 (iii) Important
Ex 2.3, 3 (iv)
Ex 2.3, 4 (i) Deleted for CBSE Board 2024 Exams
Ex 2.3, 4 (ii) Important
Ex 2.3, 4 (iii)
Ex 2.3, 4 (iv) Important
Ex 2.3, 5 (i) Deleted for CBSE Board 2024 Exams
Ex 2.3, 5 (ii) Important
Ex 2.3, 5 (iii) Important
Ex 2.3, 5 (iv) You are here
Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
-
Ex 2.3
- Ex 2.4
- Examples
- Remainder Theorem
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)
