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

Question

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?

Without foil I get $4y^2-4$ but don't understand why the answers are so different. — Liger86, Oct 31 '11 at 19:14
It's -8y. What do you mean by "solving" though? Do you mean expanding? Solve usually has to do with an equation, like solve $(2y-2)^2 = 0$. But, if you want to do that, it's already in the perfect form. You can see, without any work, that y = 1 is the only solution. — GeoffDS, Oct 31 '11 at 19:16
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. — hmakholm left over Monica, Oct 31 '11 at 19:17
@Liger86: No. $$\begin{align}(2y-2)(2y-2) &= 2y(2y-2) -2(2y-2)\ &= (2y)(2y) +(2y)(-2) +(-2)(2y) +(-2)(-2)\ &= 4y^2 -4y -4y + 4.\end{align}$$ — Arturo Magidin, Oct 31 '11 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. — Arturo Magidin, Oct 31 '11 at 19:24

AMPerrine · Accepted Answer · 2011-10-31T19:24:51.573

5

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

$$(2y)^2+2(2y)(-2)+(-2)^2=4y^2-8y+4$$

edited Oct 31 '11 at 19:24

answered Oct 31 '11 at 19:19

AMPerrine

1 Answers1