Last updated at May 29, 2018 by Teachoo

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Ex 2.4, 3 Find the value of k, if x 1 is a factor of p(x) in each of the following cases: (i) p(x) = x2 + x + k Finding remainder when x2 + x + k is divided by x 1 Step 1: Put Divisor = 0 x 1 = 0 x = 1 Step 2: Let p(x) = x2 + x + k Putting x = 1 p(1) = (1)2 + 1 + k = 1 + 1 + k = 2 + k Thus, Remainder = p(1) = 2 + k Since x 1 is a factor of x2 + x + k Remainder is zero, 2 + k = 0 k = 2 Thus, k = 2 Ex 2.4, 3 Find the value of k, if x 1 is a factor of p(x) in each of the following cases: (ii) p(x) = 2x2 + kx + 2 Finding remainder when 2x2 + kx + 2 is divided by x 1 Step 1: Put Divisor = 0 x 1 = 0 x = 1 Step 2: Let p(x) = 2x2 + kx + 2 Putting x = 1 p(1) = 2(1)2 + k + 2 = 2 + k + 2 = 2 + 2 + k Thus, Remainder = p(1) = 2 + 2 + k Since x 1 is a factor of 2x2 + kx + 2 Remainder is zero, 2 + 2 + k = 0 k = 2 2 Thus, k = 2 2 Ex 2.4, 3 Find the value of k, if x 1 is a factor of p(x) in each of the following cases: (iii) p(x) = kx2 2 x + 1 Finding remainder when kx2 2x + 1 is divided by x 1 Step 1: Put Divisor = 0 x 1 = 0 x = 1 Step 2: Let p(x) = kx2 2x + 1 Putting x = 1 p(1) = k(1)2 2 (1) + 1 = k 2 + 1 Thus, Remainder = p(1) = k 2 + 1 Since x 1 is a factor of kx2 2x + 1 Remainder is zero, k 2 + 1 = 0 k = 2 1 Thus, k = 2 1 Ex 2.4, 3 Find the value of k, if x 1 is a factor of p(x) in each of the following cases: (iv) p(x) = kx2 3x + k Finding remainder when kx2 3x + k is divided by x 1 Step 1: Put Divisor = 0 x 1 = 0 x = 1 Step 2: Let p(x) = kx2 3x + k Putting x = 1 p(1) = k(1)2 3(1) + k = k 3 + k = 2k 3 Thus, Remainder = p(1) = 2k 3 Since x 1 is a factor of x2 + x + k Remainder is zero, 2k 3 = 0 2k = 3 k = 3/2 Thus, k = 3/2

Ex 2.1, 4
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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 You are here

Ex 2.4,4 Important

Ex 2.4,5 Important

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.