Ex 8.3

Chapter 8 Class 8 Algebraic Expressions and Identities
Serial order wise

### Transcript

Ex 8.3, 5 (d) Subtract: 3a (a + b + c) – 2 b (a – b + c) from 4c(–a + b + c)Simplifying expressions 3a (a + b + c) – 2b (a – b + c) = (3a × a) + (3a × b) + (3a × c) – (2b × a) – (2b × –b) – (2b × c) = 3a2 + 3ab + 3ac – 2ba – (–2b2) – 2bc = 3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc = 3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc = 3a2 + 2b2 + (3ab – 2ab) – 2bc + 3ac = 3a2 + 2b2 + ab – 2bc + 3ac 4c (– a + b + c) = (4c × – a) + (4c × b) + (4c × c) = – 4ca + 4cb + 4c2 = 4c2 + 4bc – 4ac So, our expressions are 3a2 + 2b2 + ab – 2bc + 3ac & 4c2 + 4bc – 4ac We have to subtract 1st expression from the 2nd expression. So, we write 2nd expression first. Putting like terms below like terms So, our answer is −3a2 − 2b2 + 4c2 − ab + 6bc − 7ac

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.