# 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 $$x – 2 = sqrt{epsilon}$$