Is it wrong to factor in this way? I found an exercise for factoring:
$\frac{4x^2-8x+4}{x^2-1} * \frac{x+1}{12}$
I resolve like this:
= $\frac{(2x -2)^2}{(x-1)(x+1)} * \frac{(x+1)}{12}$
= $\frac{(2x -2)^2}{12(x-1)}$
Well up here, because here comes my problem:
$1-.$ If I factor like this:
= $\frac{2(x  -1)^2}{12(x-1)}$
= $\frac{(x-1)}{6}$ <--- First answer
$2-.$ If I factor with this other way:
= $\frac{(2x-2)(2x-2)}{12(x-1)}$
= $\frac{4(x-1)(x-1)}{12(x-1)}$
= $\frac{(x-1)}{3}$ <-- Second answer
then, what is the wrong answer ? and why ?

When you factor something out of an expression which is squared, be wary of where the brackets go
$$(2x-2)^2=[2(x-1)]^2=[2]^2[x-1]^2=4(x-1)^2$$
So the first answer is wrong, the second is correct.
