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

Transcript

Ex 14.4, 13 Find and correct the errors in the following mathematical statements. (2a + 3b) (a โ€“ b) = 2๐‘Ž^2 โ€“ 3๐‘^2 (2a + 3b) (a โ€“ b) = 2๐‘Ž^2 โ€“ 3๐‘^2 Solving L.H.S : (2a + 3b) (a โ€“ b) = 2a (a โˆ’ b) + 3b (a โˆ’ b) = (2a ร— a) โˆ’ (2a ร— b) + (3b ร— a) โˆ’ (3b ร— b) = ใ€–2๐‘Žใ€—^2 โˆ’ 2ab + 3ba โˆ’ ใ€–3๐‘ใ€—^2 = ใ€–2๐‘Žใ€—^2 โˆ’ 2ab + 3ab โˆ’ ใ€–3๐‘ใ€—^2 Taking ab common = ใ€–2๐‘Žใ€—^2 + ab (โˆ’2 + 3) โˆ’ ใ€–3๐‘ใ€—^2 (Since 3ba = 3ab) = ใ€–2๐‘Žใ€—^2 + ab โˆ’ ใ€–3๐‘ใ€—^2 โˆด (2a + 3b) (a โˆ’ b) = ใ€–๐Ÿ๐’‚ใ€—^๐Ÿ + ab โˆ’ ใ€–๐Ÿ‘๐’ƒใ€—^๐Ÿ is the correct statement

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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.