Question 410486
difference of squares:
a^2 - b^2 = (a-b) * (a+b)


equation is:
4*(x+y)^2 - (2y-z)^2


let a = 4*(x+y)^2 = 2^2*(x+y)^2 = (2*(x+y))^2 = (2x+2y)^2
let b = (2y-z)


equation becomes:
a^2 - b^2
which is the difference of squares which is equivalent to:
(2x+2y)^2 - (2y-z)^2


a^2 - b^2 = (a-b) * (a+b)


this is equivalent to:
(2x+2y)^2 - (2y-z)^2 = ((2x+2y) - (2y-z)) * ((2x+2y) + (2y-z))
because:
(2x+2y) = a
and:
(2y-z) = b


this should be the answer you are looking for:
(2x+2y)^2 - (2y-z)^2 = ((2x+2y) - (2y-z)) * ((2x+2y) + (2y-z))


you should now be able to solve as follows:
(2x+2y)^2 - (2y-z)^2 = ((2x+2y) - (2y-z)) * ((2x+2y) + (2y-z))
simplify expression on right of equal sign to get:
(2x+2y)^2 - (2y-z)^2 = (2x+2y -2y + z)) * (2x+2y + 2y-z)
simplify further to get:
(2x+2y)^2 - (2y-z)^2 = (2x+z)) * (2x+4y-z)
simplify further by multiplying expression on right of equal of sign to get:
(2x+2y)^2 - (2y-z)^2 = 4x^2 + 8xy + 4yz - z^2


this would be your answer after you've simplified as far as you can go:
(2x+2y)^2 - (2y-z)^2 = 4x^2 + 8xy + 4yz - z^2


you would confirm by substituting random numbers for x, y, and z.


I used x = 3, y = 4, z = 5 and I got:
187 = 187
this confirmed it for me.


I substituted in the original expression of:
4*(x+y)^2 - (2y-z)^2 to get 187


I then substituted in the final expression of:
4x^2 + 8xy + 4yz - z^2 to also get 187


the key was to get the original expression to look like a^2 - b^2


that was done up top when I did the following:


let a = 4*(x+y)^2 = 2^2*(x+y)^2 = (2*(x+y))^2 = (2x+2y)^2
let b = (2y-z)


the "a" part was the tricky part.


you needed to know that 4 = 2^2 and that 2^2 * (x+y)^2 = (2*(x+y)^2 = (2x+2y)^2