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  1. Chapter 14 Class 8 Factorisation
  2. Serial order wise

Transcript

Example 15 Divide 44 (๐‘ฅ^4 โ€“ 5๐‘ฅ^3 โ€“ 24๐‘ฅ^2) by 11x (x โ€“ 8) We first factorize 44 (๐‘ฅ^4 โ€“ 5๐‘ฅ^3 โ€“ 24๐‘ฅ^2) = 44 (๐‘ฅ^2 ร— ๐‘ฅ^2 โˆ’ 5๐‘ฅ ร— ๐‘ฅ^2 โˆ’ 24 ร— ๐‘ฅ^2) Taking ๐‘ฅ^2 common, = ใ€–44๐‘ฅใ€—^2 (๐‘ฅ^2 โˆ’ 5๐‘ฅ โˆ’ 24) By middle term splitting, = 44๐‘ฅ^2 (๐‘ฅ^2+3๐‘ฅ โˆ’ 8๐‘ฅ โˆ’ 24) = 44๐‘ฅ^2 [(๐‘ฅ^2+3๐‘ฅ) โˆ’ (8๐‘ฅ+24)] Both have x as common factor Both have 8 as common factor Splitting the middle term We need to find two numbers whose Sum = โˆ’5 Product = โˆ’24 So, we write โˆ’5x = 3x โˆ’ 8x = 44๐‘ฅ^2 [๐‘ฅ(๐‘ฅ+3) โˆ’8(๐‘ฅ+3)] Taking ๐‘ฅ+3 common, = 44๐‘ฅ^2 (๐‘ฅ+3) (๐‘ฅ โˆ’ 8) Dividing (44 (๐‘ฅ^4 โˆ’ 5๐‘ฅ^3 โˆ’ 24๐‘ฅ^2 ))/(11๐‘ฅ (๐‘ฅ โˆ’ 8)) = (44 ๐‘ฅ^2 (๐‘ฅ + 3) (๐‘ฅ โˆ’ 8))/(11๐‘ฅ (๐‘ฅ โˆ’ 8)) = 44/11 ร— ๐‘ฅ^2/๐‘ฅ ร— (๐‘ฅ + 3) ร— ((๐‘ฅ โˆ’ 8))/((๐‘ฅ โˆ’ 8)) = 4 ร— ๐‘ฅ ร— (๐‘ฅ + 3) = 4๐’™ (๐’™ + 3)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.