Ex 2.4,3 - Chapter 2 Class 9 Polynomials - Part 7

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Ex 2.4,3 - Chapter 2 Class 9 Polynomials - Part 8

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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: (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

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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.