1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise
  3. Ex 2.4

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

Last updated at Aug. 25, 2021 by

Ex 2.4,3 - Chapter 2 Class 9 Polynomials - Part 5

Ex 2.4,3 - Chapter 2 Class 9 Polynomials - Part 6

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

    3. Ex 2.4

    Ex 2.5 →

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: (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

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

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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.