Ex 8.4, 3 Simplify. (iv) (𝑎+𝑏) (𝑐−𝑑)+(𝑎−𝑏) (𝑐+𝑑)+2 (𝑎𝑐+𝑏𝑑)Here, there are 3 expressions 1st expression = (𝑎+𝑏) (𝑐−𝑑) 2nd expression = (𝑎−𝑏) (𝑐+𝑑) 3rd expression = 2 (𝑎𝑐+𝑏𝑑) Solving 1st expression (𝒂+𝒃) (𝒄−𝒅) = 𝑎(𝑐−𝑑)+𝑏 (𝑐−𝑑) = (𝑎 × 𝑐)−(𝑎 × 𝑑)+(𝑏 × 𝑐)−(𝑏 × 𝑑) = 𝒂𝒄−𝒂𝒅+𝒃𝒄−𝒃𝒅 Solving 2nd expression (𝒂−𝒃) (𝒄+𝒅) = 𝑎(𝑐+𝑑)−𝑏 (𝑐+𝑑) = (𝑎×𝑐)+(𝑎×𝑑)−(𝑏×𝑐)−(𝑏−𝑑) = 𝒂𝒄+𝒂𝒅−𝒃𝒄−𝒃𝒅 Solving 3rd expression 2 (𝑎𝑐+𝑏𝑑) = (2×𝑎𝑐)+(2×𝑏𝑑) = 𝟐𝒂𝒄+𝟐𝒃𝒅 Now, our expression becomes (𝒂+𝒃) (𝒄−𝒅)+(𝒂−𝒃) (𝒄+𝒅)+𝟐 (𝒂𝒄+𝒃𝒅) = 𝑎𝑐−𝑎𝑑+𝑏𝑐−𝑏𝑑+𝑎𝑐+𝑎𝑑−𝑏𝑐−𝑏𝑑+2𝑎𝑐+2𝑏𝑑 = (𝑎𝑐+𝑎𝑐+2𝑎𝑐)+(−𝑎𝑑+𝑎𝑑)+(𝑏𝑐−𝑏𝑐)+(−𝑏𝑑−𝑏𝑑+2𝑏𝑑) = (1+1+2)𝑎𝑐+(−1+1)𝑎𝑑+(1−1)𝑏𝑐+(−1−1+2)𝑏𝑑 = 4𝑎𝑐+(0)𝑎𝑑+(0)𝑏𝑐+(0)𝑏𝑑 = 𝟒𝒂𝒄

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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.