Example 14 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2

Example 14 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3

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Example 14 Use the Identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following: (ii) 95 Γ— 103 95 Γ— 103 = (100βˆ’5)Γ—(100+3) = [100+(βˆ’5)]Γ—[100+3] (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘₯ = 100 , π‘Ž = βˆ’5 & 𝑏 = 3 = (100)^2+(βˆ’5+2)(100)+(βˆ’5)(3) = 10000+(βˆ’2Γ—100)+(βˆ’15) = 10000βˆ’200βˆ’15 = 1000βˆ’215 = πŸ—πŸ•πŸ–πŸ“

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