Last updated at March 16, 2017 by Teachoo

Transcript

Ex 2.4, 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) 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, 5 Factorise: (iii) x3 + 13x2 + 32x + 20 Let p(x) = x3 + 13x2 + 32x + 20 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 + 12x + 20 So, p(x) = (x + 1) g(x) = (x + 1) (x2 + 12x + 20) We factorize g(x) i.e. x2 + 12x + 20 x2 + 12x + 20 We factorize using the splitting the middle term method = x2 + 10x + 2x + 20 = x(x + 10) + 2(x + 10) = (x + 10) (x + 2) So, p(x) = (x + 1)(x + 2)(x + 10) Ex 2.4, 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)

Chapter 2 Class 9 Polynomials

Ex 2.1, 4
Important

Example 3 Important

Ex 2.2,1 Important

Ex 2.2,3 Important

Ex 2.2,4 Important

Example 7 Important

Ex 2.3,2 Important

Ex 2.3,3 Important

Example 12 Important

Example 15 Important

Ex 2.4,1 Important

Ex 2.4,2 Important

Ex 2.4,3 Important

Ex 2.4,4 Important

Ex 2.4,5 Important You are here

Example 18 Important

Example 21 Important

Example 23 Important

Example 24 Important

Ex 2.5,2 Important

Ex 2.5,3 Important

Ex 2.5,4 Important

Ex 2.5,5 Important

Ex 2.5,7 Important

Ex 2.5,10 Important

Ex 2.5, 11 Important

Ex 2.5,16 Important

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

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.