SOLUTION: Hi, Please help me solve this math problem... Directions are :factor each expression as a difference of squares. 4(x+y)^2 -(2y-z)^2 my teacher told us to substitute, so i di

Question 410486: Hi, Please help me solve this math problem...
Directions are :factor each expression as a difference of squares.
4(x+y)^2 -(2y-z)^2
my teacher told us to substitute, so i did : a= xty,b= 2y-z
4(a)^2-(b)^2
well i'm not sure i got the right answer and i'm still not 100% sure on what to do so if you could please try to explain it to me I would greatly appreciate it, Thank you XD
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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

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