Ex 2.3, 5 (i) - Chapter 2 Class 9 Polynomials (Important Question)
Last updated at Dec. 13, 2024 by Teachoo
Chapter 2 Class 9 Polynomials
Ex 2.1, 4 (i)
Important
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Class 9
Important Questions for Exam - Class 9
- Chapter 1 Class 9 Number Systems
-
Chapter 2 Class 9 Polynomials
- Chapter 3 Class 9 Coordinate Geometry
- Chapter 4 Class 9 Linear Equations in Two Variables
- Chapter 5 Class 9 Introduction to Euclid's Geometry
- Chapter 6 Class 9 Lines and Angles
- Chapter 7 Class 9 Triangles
- Chapter 8 Class 9 Quadrilaterals
- Chapter 9 Class 9 Areas of parallelograms and Triangles
- Chapter 10 Class 9 Circles
- Chapter 11 Class 9 Constructions
- Chapter 12 Class 9 Herons Formula
- Chapter 13 Class 9 Surface Areas and Volumes
- Chapter 14 Class 9 Statistics
- Chapter 15 Class 9 Probability
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)
