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Ex 2.5, 5 - Factorise (i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - Ex 2.5

  1. Chapter 2 Class 9 Polynomials
  2. Serial order wise
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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" )

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