Where's Waldo?

Question

How in the hell do I factor: (show work + explaination)

9y²-(2x-y)²=

Answer 1

9y^2-(2x-y)^2

9y^2-4x^2+4xy-y^2

8y^2-4x^2+4xy

Assume x=5

8y^2-100+20y

8y^2+40y-20y-100

8y(y+5)-20(y+5)

(8y-20)(y+5)

5=x

(8y-4x)(y+x)

I chose a number unrelated to ther other numbers to make sure I can factor out 5 again.

Answer 2

Hello

This is the difference between squares so we have

[3y -(2x-y)]*[3y+(2x-y)]

3y -(2x -y) = 4y-2x and 3y+2x-y = 2y+2x so we have
(4y-2x)(2y+2x) Taking out a 2 from the first binomial we have

2(2y-x) and a 2 from the second gives us 2(y+x) or

4(2y-x)(y+x)

Hope This Helps!!

Answer 3

it is the difference of two squares

9y² - (2x - y)² = (3y)² - (2x - y)²
= [(3y) + (2x - y)][(3y) - (2x - y)]
= [3y + 2x - y][3y - 2x + y]
= (2y + 2x )(4y - 2x)
=2(y + x) 2 (2y - x)
=4(y + x)(2y - x)