Algebra Help Please!! Exam Tomorrow

Hey guys, im having a little bit of trouble with the following question:
1-(2y-3)^2

Basically what im doing is going as follows
[1-(2y-3)][1+(2y-3)]
(4-2y)(2y-2)
2(2-y)(y-1)
But apparently thats wrong, WHY?!?!?

Thanks in advance

This was so smart to factor as difference of two squares!
Your only mistake is treating factors like you do when you distribute.
(4-2y)(2y-2)
Factor 2 out of EACH factor
2(2-y)(2)(y-1)
4(2-y)(y-1)
Another thing is the second factor is not in proper form, so we should also factor out -1
So
-4(y-2)(y-1)

If you didn't see the a^2-b^2 at the beginning, you could square the binomial, then combine with the 1 and re-factor
Good work!

well 1 is a square but it's still different than the difference of two squares because for that the whole equation is being squared after the subtraction... for example 1^2- 2^2 = -1 but (1-2)^2= 1

so what you want to do is use the difference of two squares one the second part

1- (2y+3) (2y-3)

and then solve this


hope i helped!

The way you have it written, only the 2y-3 is squared.

1 - (2y-3)^2
1 - (4y^2 - 12y + 9)
1 - 4y^2 + 12y - 9
-4y^2 + 12y - 8

1 - (2y-3)2
1 - (2y-3)(2y-3)
1 - 4y2 - 12y + 9
-4y2 - 12y + 10

That's the right answer, 100%. Haters gonna hate :P

Just the last term (2y-3) is squared the way it is written...

1 - 4y^2 - 6y + 9 = 4y^2 - 6y + 10

1-(8/y^6)=(y^6-8)/y^6 is O.K.
god bless you.