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

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Ex 2.3,1 - Chapter 2 Class 9 Polynomials - Part 10

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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 (v) 5 + 2x Dividing x3 + 3x2 + 3x + 1 by 5 + 2π‘₯ Step 1: Put Divisor = 0 5 + 2x = 0 2x = –5 x = – 5/2 Step 2: Let p(x) = x3 + 3x2 + 3x + 1 Putting x = – 5/2 p((βˆ’5)/2) = ((βˆ’5)/2)^3+ 3((βˆ’5)/2)^2 + 3((βˆ’5)/2) + 1 = (βˆ’125)/8 + 3(25/4) – 15/2 + 1 = (βˆ’125)/8 + 75/4 – 15/2 + 1 = (βˆ’125 + 75(2) βˆ’ 15(4) + 1(8) )/8 = (βˆ’125 + 150 βˆ’ 60 + 8)/8 = (βˆ’27)/8 Thus, remainder = p((βˆ’5)/2) = (βˆ’27)/8

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