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Ex 2.3, 5 (i) - 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 You are here
Ex 2.3, 5 (ii) Important
Ex 2.3, 5 (iii) Important
Ex 2.3, 5 (iv)
Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
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Ex 2.3
- Ex 2.4
- Examples
- Remainder Theorem
Transcript
Ex 2.3, 5 Factorise: (i) x3 − 2x2 − x + 2 Let p(x) = x3 – 2x2 – x + 2 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 – x – 2 So, p(x) = (x – 1) g(x) = (x – 1) (x2 – x – 2) We factorize g(x) i.e. x2 – x – 2 x2 – x – 2 We factorize using the splitting the middle term method = x2 – 2x + x – 2 = x(x – 2) + 1 (x – 2) = (x + 1) (x – 2) So, p(x) = (x – 1)(x + 1)(x – 2)
