Ex 2.3, 3 (ii) - 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: (ii) p(x) = 2x2 + kx + 2 Finding remainder when 2x2 + kx + 2 is divided by x 1 Step 1: Put Divisor = 0 x 1 = 0 x = 1 Step 2: Let p(x) = 2x2 + kx + 2 Putting x = 1 p(1) = 2(1)2 + k + 2 = 2 + k + 2 = 2 + 2 + k Thus, Remainder = p(1) = 2 + 2 + k Since x 1 is a factor of 2x2 + kx + 2 Remainder is zero, 2 + 2 + k = 0 k = 2 2 Thus, k = 2 2
