Ex 2.4, 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" )

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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