Ex 2.4, 3 - Find value of k, if x - 1 is a factor of p(x) - Ex 2.4

 

  1. Chapter 2 Class 9 Polynomials
  2. Serial order wise
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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: (i) p(x) = x2 + x + k Finding remainder when x2 + x + k is divided by x 1 Step 1: Put Divisor = 0 x 1 = 0 x = 1 Step 2: Let p(x) = x2 + x + k Putting x = 1 p(1) = (1)2 + 1 + k = 1 + 1 + k = 2 + k Thus, Remainder = p(1) = 2 + k Since x 1 is a factor of x2 + x + k Remainder is zero, 2 + k = 0 k = 2 Thus, k = 2 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 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 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.