Ex 2.4, 5 (ii) - Chapter 2 Class 9 Polynomials
Last updated at Dec. 16, 2024 by Teachoo
Ex 2.4
Ex 2.4, 1 (i)
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Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
- Ex 2.3
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Ex 2.4
- Examples
- Remainder Theorem
Transcript
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")
