Ex 2.3, 3 (iii) - Chapter 2 Class 9 Polynomials
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 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: (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 2.3
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)
Ex 2.3, 3 (ii) Important
Ex 2.3, 3 (iii) Important You are here
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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