Ex 2.4,3 - Chapter 2 Class 9 Polynomials - Part 3

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Ex 2.4,3 - Chapter 2 Class 9 Polynomials - Part 4

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  1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise

Transcript

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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.