
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Examples
Example 2 Important
Example 3
Example 4 Important
Example 5 (i)
Example 5 (ii) Important
Example 6 (i)
Example 6 (ii) Important You are here
Example 7
Example 8 (i)
Example 8 (ii)
Example 9 (i)
Example 9 (ii) Important
Example 10
Question 1 (i) Deleted for CBSE Board 2024 Exams
Question 1 (ii) Important Deleted for CBSE Board 2024 Exams
Question 2 (i) Deleted for CBSE Board 2024 Exams
Question 2 (ii) Important Deleted for CBSE Board 2024 Exams
Question 3 (i) Deleted for CBSE Board 2024 Exams
Question 3 (ii) Important Deleted for CBSE Board 2024 Exams
Question 3 (iii) Deleted for CBSE Board 2024 Exams
Question 4 (i) Deleted for CBSE Board 2024 Exams
Question 4 (ii) Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 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