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Ex 2.4, 3 (iii) - Chapter 2 Class 9 Polynomials (Term 2)

Last updated at Aug. 25, 2021 by

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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: (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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.