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Question

Factorise:
2x2+y2+8z222xy+42yz8xz   [3 MARKS]


Solution

Formula: 1 Mark
Factorisation: 2 Marks

2x2+y2+8z222xy+42yz8xz

=(2x)2+y2+(22z)22(2x)(y)+2(y)(22z)2(2x)(22z)

=(2x)2+y2+(22z)2+2(2x)(y)+2(y)(22z)+2(2x)(22z).....(i)


(x+y+z)2=x2+y2+z2+2xy+2yz+2zx
[Comparing equation (i) from above identity, we get]

=(2x+y+22z)2

=(2x+y+22z)(2x+y+22z)

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