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Ex 2.3, 1 (v) - Chapter 2 Class 9 Polynomials (Term 2)

Last updated at March 22, 2023 by

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

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

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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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.