Ex 2.3,1 - Chapter 2 Class 9 Polynomials - Part 7

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Ex 2.3,1 - Chapter 2 Class 9 Polynomials - Part 8

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  1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise

Transcript

Ex 2.3, 1 Find the remainder when x3 + 3x2 + 3x + 1 is divided by (iv) x + π Dividing x3 + 3x2 + 3x + 1 by x + π Step 1: Put Divisor = 0 x + π = 0 x = – π Step 2: Let p(x) = x3 + 3x2 + 3x + 1 Putting x = – π p("−" 𝜋) = ("−" 𝜋)3 + 3("−" 𝜋)2 + 3("−" 𝜋) + 1 = – 𝜋3 + 3𝜋2 – 3𝜋 + 1 Thus, remainder = p("−" 𝜋) = – 𝜋3 + 3𝜋2 – 3𝜋 + 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.