Last updated at May 29, 2018 by Teachoo

Transcript

Ex 2.5, 5 Factorise: 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz = 22 x2 + 32 y2 + 42 z2 + 12xy – 24yz – 16xz = (2x)2 + (3y)2 + (4z)2 + 12xy – 24yz – 16xz = (2x)2 + (3y)2 + (–4z)2 + 12xy – 24yz – 16xz = (2x)2 + (3y)2 + (–4z)2 + 2(2x)(3y) + 2(3y) (–4z) + 2(2x) (–4z) Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac Putting a = 2x, b= 3y , c = –4z = (2x + 3y – 4z)2 = (2x + 3y – 4z) (2x + 3y – 4z) Ex 2.5, 5 Factorise: (ii) 2x2 + y2 + 8z2 – 2√2xy + 4 √2 yz – 8xz 2x2 + y2 + 8z2 – 2√2xy + 4√2 yz – 8xz = (√2 )^2x2 + y2 + (2√2 )^2z2 – 2√2xy + 4√2 yz – 8xz = (√2 𝑥)^2 + y2 + (2√2 𝑧)^2 – 2√2xy + 4√2 yz – 8xz = (−√2 𝑥)^2 + y2 + (2√2 𝑧)^2 – 2√2xy + 4√2 yz – 8xz = (−√2 𝑥)^2+ y2 + (2√2 𝑧)^2+ 2(–√2x)(y) + 2(y)(2√2z) + 2(–√2x)(2√2z) Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac Putting a = -√2x, b= y , c = 2√2z = ("–" √2 "x + y + 2 " √2 "z" )^2 = ("–" √2 "x + y + 2 " √2 "z" )("–" √2 "x + y + 2 " √2 "z" )

Ex 2.5

Ex 2.5, 1

Ex 2.5,2 Important

Ex 2.5,3 Important

Ex 2.5,4 Important

Ex 2.5,5 Important You are here

Ex 2.5,6

Ex 2.5,7 Important

Ex 2.5,8

Ex 2.5,9

Ex 2.5,10 Important

Ex 2.5, 11 Important

Ex 2.5,12 Deleted for CBSE Board 2022 Exams

Ex 2.5,13 Deleted for CBSE Board 2022 Exams

Ex 2.5,14 Deleted for CBSE Board 2022 Exams

Ex 2.5, 15

Ex 2.5,16 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.