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 is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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