# Expanding $(2y-2)^2$ by FOIL

Expanding $$(2y-2)^2$$

Isn't this same as $$\begin{gather*} (2y-2)(2y-2)\\ = 4y^2-6y+4\ ? \end{gather*}$$

This should be FOIL, shouldn't it?

• Where on earth do you get 6 from?? Commented Oct 31, 2011 at 19:15
• And how do you get $4y^2-4$? Show some intermediate results so we can see what goes wrong. It ought to be OK to ditch "FOIL" (stupid rule, addition is commutative so there is no point going around remembering a particular order of the terms -- it doesn't matter whether you do FOIL or IOLF or FLOI or FILO) -- but you still have to apply the distributive rule correctly. Commented Oct 31, 2011 at 19:17
• @Liger86: No. \begin{align*}(2y-2)(2y-2) &= 2y(2y-2) -2(2y-2)\\ &= (2y)(2y) +(2y)(-2) +(-2)(2y) +(-2)(-2)\\ &= 4y^2 -4y -4y + 4.\end{align*} Commented Oct 31, 2011 at 19:23
• Grrr... "be FOILd". I'd like to join the club and beat whoever came up with that idiotic acronym over the head with it. Commented Oct 31, 2011 at 19:24
• What on earth is FOIL??? Commented Oct 31, 2011 at 20:39

Yes, $(2y-2)^2=(2y-2)(2y-2)$.
FOILing should work, but will get you $4y^2−8y+4$, rather than $4y^2−6y+4$, as shown:
$(2y-2)(2y-2)=(2y)(2y)+(2y)(-2)+(-2)(2y)+(-2)(-2)$ $=4y^2-4y-4y+4=4y^2−8y+4$
If you want to memorize the formula which will get you the same result, it is $$(a+b)^2=a^2+2ab+b^2$$ In your example, $a=2y$ and $b=-2$, so you get
$$(2y)^2+2(2y)(-2)+(-2)^2=4y^2-8y+4$$