# Ex 14.3, 3 (iii) - Chapter 14 Class 8 Factorisation

Last updated at Dec. 26, 2018 by Teachoo

Last updated at Dec. 26, 2018 by Teachoo

Transcript

Ex 14.3, 3 Work out the following divisions. (iii) 10y(6y + 21) ÷ 5(2y + 7) We first factorise 10y (6y + 21) = 10y [(3 × 2) y + (3 × 7)] Taking 3 common, = 10y × 3 (2y + 7) Dividing, (10𝑦 (6𝑦 + 21) )/(5 (2𝑦 +7)) = (10𝑦 × 3(2𝑦 + 7))/(5 × (2𝑦 + 7)) = 3 × 10/5 × y × ((2𝑦 + 7))/((2𝑦 + 7)) = 3 × 2 × y × 1 = 6y

Ex 14.3

Ex 14.3, 1 (i)

Ex 14.3, 1 (ii)

Ex 14.3, 1 (iii)

Ex 14.3, 1 (iv)

Ex 14.3, 1 (v) Important

Ex 14.3, 2 (i)

Ex 14.3, 2 (ii) Important

Ex 14.3, 2 (iii) Important

Ex 14.3, 2 (iv)

Ex 14.3, 2 (v)

Ex 14.3, 3 (i)

Ex 14.3, 3 (ii)

Ex 14.3, 3 (iii) Important You are here

Ex 14.3, 3 (iv)

Ex 14.3, 3 (v) Important

Ex 14.3, 4 (i)

Ex 14.3, 4 (ii)

Ex 14.3, 4 (iii) Important

Ex 14.3, 4 (iv)

Ex 14.3, 4 (v)

Ex 14.3, 5 (i)

Ex 14.3, 5 (ii) Important

Ex 14.3, 5 (iii) Important

Ex 14.3, 5 (iv) Important

Ex 14.3, 5 (v)

Ex 14.3, 5 (vi)

Ex 14.3, 5 (vii) Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.