Ex 2.3, 3 (iii) - Chapter 2 Class 9 Polynomials

Last updated at Dec. 16, 2024 by

Ex 2.3, 3 - Find k, if (x - 1) is a factor of p(x) = kx^2 - √2x + 1 - Ex 2.3

part 2 - Ex 2.3, 3 (iii) - Ex 2.3 - Serial order wise - Chapter 2 Class 9 Polynomials

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Ex 2.3, 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo