2
$\begingroup$

Expanding $(2y-2)^2$

Isn't this same as $$(2y-2)(2y-2)\ ?$$

$$4y^2-6y+4$$

This should be FOILd shouldn't it?

  • 0
    Without foil I get $4y^2-4$ but don't understand why the answers are so different.2011-10-31
  • 3
    Where on earth do you get 6 from??2011-10-31
  • 0
    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.2011-10-31
  • 4
    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.2011-10-31
  • 0
    $4y^2-2y-4y+4$ right?2011-10-31
  • 0
    Um, does that mean that you get $-2y$ from $2y\times -2$?2011-10-31
  • 1
    @Liger86: No. $$\begin{align*}(2y-2)(2y-2) &= 2y(2y-2) -2(2y-2)\\ &= (2y)(2y) +(2y)(-2) +(-2)(2y) +(-2)(-2)\\ &= 4y^2 -4y -4y + 4.\end{align*}$$2011-10-31
  • 5
    Grrr... "be FOILd". I'd like to join the club and beat whoever came up with that idiotic acronym over the head with it.2011-10-31
  • 3
    What on earth is FOIL???2011-10-31

1 Answers 1

4

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

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