Example 6 (ii) - Chapter 8 Class 8 Algebraic Expressions and Identities
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
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
Examples
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Example 4 Important
Example 5 (i)
Example 5 (ii) Important
Example 6 (i)
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Example 7
Example 8 (i)
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Example 10
Question 1 (i)
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Question 2 (i)
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Question 3 (iii)
Question 4 (i)
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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