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Ex 2.3, 3 (i) - Chapter 2 Class 9 Polynomials

Last updated at May 29, 2023 by

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

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

 

 

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Ex 2.3, 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

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