Ex 9.5, 1 - Use a suitable identity to get product (i) (x + 3) (x + 3)

Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3 Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4 Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5 Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 6 Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 7 Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 8 Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 9 Ex 9.5, 1 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 10

  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

Ex 9.5, 1 Use a suitable identity to get each of the following products. (i) (๐‘ฅ+3) (๐‘ฅ+3) (๐‘ฅ+3) (๐‘ฅ+3) = (๐‘ฅ+3)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = ๐‘ฅ & ๐‘ = 3 = ๐‘ฅ^2+3^2+2(๐‘ฅ) (3) = ๐’™^๐Ÿ+๐Ÿ—+๐Ÿ”๐’™ Ex 9.5, 1 Use a suitable identity to get each of the following products. (ii) (2๐‘ฆ + 5) (2๐‘ฆ + 5) (2๐‘ฆ+5) (2๐‘ฆ+5) = (2๐‘ฆ+5)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 2๐‘ฆ & ๐‘ = 5 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 2๐‘ฆ & ๐‘ = 5

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.