Ex 2.3, 3 (i) - Chapter 2 Class 9 Polynomials
Last updated at Dec. 13, 2024 by Teachoo
Last updated at Dec. 13, 2024 by Teachoo
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
Ex 2.3
Ex 2.3, 1 (ii)
Ex 2.3, 1 (iii)
Ex 2.3, 1 (iv) Important
Ex 2.3, 2 (i)
Ex 2.3, 2 (ii) Important
Ex 2.3, 2 (iii)
Ex 2.3, 3 (i) You are here
Ex 2.3, 3 (ii) Important
Ex 2.3, 3 (iii) Important
Ex 2.3, 3 (iv)
Ex 2.3, 4 (i)
Ex 2.3, 4 (ii) Important
Ex 2.3, 4 (iii)
Ex 2.3, 4 (iv) Important
Ex 2.3, 5 (i)
Ex 2.3, 5 (ii) Important
Ex 2.3, 5 (iii) Important
Ex 2.3, 5 (iv)
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