calculus – Absolute value of difference of squares

I am working through the Stewart Calculus text independently and am stuck on one of the practice problems (edition 7e – problem 29 in section 1.7).

I am confused by particularly by hint #3 which says:

Why is $lvert x^2 -4x + 4x rvert = lvert x + 2 rvert lvert x – 2 rvert$

I am quite confused by this and I cannot figure out how these two sides of the equation are equal. For full context of the homework hints please see here (I would have attached the image but I cannot since I have not posted enough).

Left to my own device I would have figured that $lvert x^2 -4x + 4rvert lt epsilon $ where the left hand sides could be reworked as $ lvert (x -2)^2 rvert lt epsilon $ and furthermore that $ x – 2 = sqrt{epsilon}$

Not sure where I am going wrong here, can somebody please help?