Example 6 (ii) - Chapter 8 Class 8 Algebraic Expressions and Identities

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Example 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3

Example 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4

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Example 6 (ii) 4y (3y2 + 5y – 7) and 2 (y3 – 4y2 + 5) Given expressions 4y (3y2 + 5y – 7) 2 (y3 – 4y2 + 5) Simplifying expressions: 4y (3y2 + 5y – 7) = (4y × 3y2 ) + (4y × 5y) + (4y × – 7) = (4 × 3 × y3 ) + (4 × 5 × y2) + (– 28 × y) = 12y3 + 20y2 – 28y 2 (y3 – 4y2 + 5) = (2 × y3) + (2 × – 4 × y2) + (2 × 5) = 2y3 – 8y2 + 10 So, our expression are 12 y3 + 20y2 – 28y & 2y3 – 8y2 + 10 Adding expressions ∴ Required sum = 14y3 + 12y2 – 28y + 10

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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 and Computer Science at Teachoo