Is it wrong to factor in this way?

Question

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 ?

Answer 1

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.

Answer 2

The first way is incorrect. You can't just factor a $2$ out of the squared expression just because you see the terms have one factor as $2$.

$$(2x-2)^2=\left[(2)(x-1)\right]^2=(2)^2(x-1)^2=4(x-1)^2\ne2(x-1)^2$$