Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 14.3, 5 (iv) - Chapter 14 Class 8 Factorisation
Last updated at March 31, 2023 by Teachoo
Ex 14.3
Ex 14.3, 1 (i)
Ex 14.3, 1 (ii) Important
Ex 14.3, 1 (iii)
Ex 14.3, 1 (iv) Important
Ex 14.3, 1 (v)
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) Important
Ex 14.3, 3 (i)
Ex 14.3, 3 (ii)
Ex 14.3, 3 (iii) Important
Ex 14.3, 3 (iv)
Ex 14.3, 3 (v) Important
Ex 14.3, 4 (i)
Ex 14.3, 4 (ii) Important
Ex 14.3, 4 (iii)
Ex 14.3, 4 (iv)
Ex 14.3, 4 (v) Important
Ex 14.3, 5 (i)
Ex 14.3, 5 (ii) Important
Ex 14.3, 5 (iii)
Ex 14.3, 5 (iv) Important You are here
Ex 14.3, 5 (v)
Ex 14.3, 5 (vi)
Ex 14.3, 5 (vii) Important
Transcript
Ex 14.3, 5 Factorise the expressions and divide them as directed. (iv) 4yz(๐ง^2 + 6z โ 16) รท 2y(z + 8) We first factorise 4yz (๐ง^2+6๐งโ16) By middle term splitting, = 4yz (๐ง^2โ2๐ง+8๐งโ16) = 4yz [(๐ง^2โ2๐ง)+(8๐งโ16)] = 4yz [z (z โ 2) + 8 (z โ 2)] Taking (z โ 2) common, = 4yz (z โ 2) (z + 8) Splitting the middle term We need to find two numbers whose Sum = 6 Product = โ16 So, we write 6z = โ2z + 8z Dividing 4yz (๐ง^2+6๐งโ16)รท2๐ฆ (๐ง+8) = (4๐ฆ๐ง (๐ง^2 + 6๐ง โ 16))/(2๐ฆ (๐ง + 8)) = (4๐ฆ๐ง (๐ง โ2) (๐ง + 8))/(2๐ฆ (๐ง + 8)) = 4/2 ร ๐ฆ/๐ฆ ร z ร (z โ 2) ร ((๐ง + 8))/((๐ง + 8)) = 2 ร z ร (z โ 2) = 2z (๐ โ 2)
