SOLUTION: factorise x^2+4y^216z^2+4xy+16yz+8zx and find the value when x=2, y=-1 and z=1

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Question 1120242: factorise x^2+4y^216z^2+4xy+16yz+8zx and find the value when x=2, y=-1 and z=1
Found 2 solutions by ikleyn, Shin123:
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
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factorise x^2+4y^216z^2+4xy+16yz+8zx and find the value when x=2, y=-1 and z=1
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First,  your expression is written  INCORRECTLY.

The correct form is  THIS: 

    factorize  x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx and find the value when x=2, y=-1 and z=1

Then 

    x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx = %28x+%2B+2y+%2B+4z%29%5E2,


and it answers the first question.


To answer the second questions, substitute the values into the right side of the formula. You will get


    %28x+%2B+2y+%2B+4z%29%5E2 = %282+-+2%2A1+%2B+4%2A1%29%5E2 = 4%5E2 = 16.

Solved.


Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
x2+4y2+16z2+4xy+16yz+8zx. First part, the equation factors to (x+2y+4z)2 Next part, (x+2y+4z)2= (2+2(-1)+4(1))2=(2+(-2)+4)2=42=16.