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Question

Factorise:
(i) 4x2+9y2+16z2+12xy24yz16xz
(ii) 2x2+yz+8z222xy+42yz8xz

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Solution

We know that, the algebraic identity:
(a2+b2+c2+2ab+2bc+2ca)=(a+b+c)2

(i)
4x2+9y2+16z2+12xy24yz16xz

Here, yz and xz terms are negative.

i.e., common variable z is in negative form.

=(2x)2+(3y)2+(4z)2+2(2x)(3y)+2(3y)(4z)+2(4z)(2x)

=(2x+3y4z)2 [(a2+b2+c2+2ab+2bc+2ca)=(a+b+c)2]

=(2x+3y4z)(2x+3y4z)

(ii)
2x2+y2+8z222xy+42yz8zx

Here, xy and zx terms are negative.

i.e., common variable x is in negative form.

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


=(2x+y+22z)2 [(a2+b2+c2+2ab+2bc+2ca)=(a+b+c)2]


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

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