Ex 2.3, 3 (iii) - Chapter 2 Class 9 Polynomials
Last updated at Dec. 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: (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
