how do you factor 4x^2-4 completely

Question

callisto · Answer

4x^2-4
= 4(x^2-1)
=4(x+1)(x-1)

tommo_lcfc · Answer

Take out a common factor of 4.

4(x^2 -1).

Recognise that (x^2-1) is a difference of two squares.

4(x-1)(x+1).

anonymous · Answer

thank you both lol

callisto · Answer

welcome

tommo_lcfc · Answer

I explain things, @Callisto doesn't! That was a compliment to me, not an insult to her!

tommo_lcfc · Answer

Anyway, you can come to us any time. I think that me and @Callisto are a team now!

callisto · Answer

sorry, sometimes workings mean explanations to me. Just ask if you don't understand. people here are willing to guide you step by step :)

anonymous · Answer

lol well thank you and im gonna need it im sure lol math is my worst subject lol its much appreciated :)

callisto · Answer

math was also my worst subject :P

anonymous · Answer

lol well looks like you got it all figured ut now lol

tommo_lcfc · Answer

I love maths! I am at university doing straight maths :)

anonymous · Answer

lol right on, i wouldnt mind it if i actually knew what i was doin :)

anonymous · Answer

how would i factor 4x^2-4x+2xy-2...

callisto · Answer

take out the common factor
4x^2-4x+2xy-2
= 2(2x^2 -2x + xy-1)

anonymous · Answer

alright

callisto · Answer

because there is a y in the third term, you cannot take out common factor and group that again.. :(

tommo_lcfc · Answer

Could you take 2x out of the first two terms and get 2(2x(x-1)+xy-1) or is that not allowed?

tommo_lcfc · Answer

It can't be factorised any further after that.

callisto · Answer

there is no use to do so, i think?

tommo_lcfc · Answer

Probably not, if you say so.

anonymous · Answer

lol well either way the answer worked, so you were right lol :) thanks.

callisto · Answer

welcome :)