Ex 2.3, 3 (iv) - Chapter 2 Class 9 Polynomials
Last updated at April 16, 2024 by Teachoo
Ex 2.3
Ex 2.3, 1 (i)
Important
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Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
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Ex 2.3
- Ex 2.4
- Examples
- Remainder Theorem
Transcript
Ex 2.3, 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
