Factor the Expression completely$ (a+b)^2 - (a-b)^2$ I don't understand this question. The answer in the book is $4ab$, but how is that term a factor?
I was thinking along the line that this was a difference of squares example. $a^2-b^2 = (a+b)(a-b)$
My answer is $[(a+b)-(a-b)][(a+b)+(a-b)]$
What do I not understand?  
 A: Just See this,
$(a+b)^2 = a^2 + b^2 +2ab$,
$(a-b)^2 = a^2 + b^2 -2ab$
Subtract them,
 You will get $4ab$
Ok I will show that , you have already done everything
$(a+b)^2 - (a-b)^2 = [(a+b)-(a-b)][(a+b)+(a-b)]$
$\hspace{2cm} = [a+b-a+b][a+b+a-b]$
$\hspace{2cm} = [a-a+b+b][a+a+b-b]$
$\hspace{2cm} = [2b][2a]$
$\hspace{2cm} = 4ab$
A: $$(a + b)^2 - (a-b)^2 = (a+b)(a+b) - (a+b)(a-b) $$ $$= (a+b)((a+b) - (a - b)) $$ $$ = (a + b)(a + b - a + b) $$ $$= (a+b)(2b)$$
POST EDIT: $$(a+b)^2 - (a-b)^2 = a^2 +2ab + b^2 -(a^2 - 2ab + b^2) = 2ab - (-2ab) = 4ab$$
A: Note that you don't even need to factor it, just expand the expression.
$$(a+b)^2-(a-b)^2$$
$$=(a^2+b^2+2ab)-(a^2+b^2-2ab)$$
$$=a^2+b^2+2ab-a^2-b^2+2ab$$
$$=4ab$$
If you have to factor it, remember the difference of squares $a^2-b^2=(a+b)(a-b)$. In this case, $a^2$ is $(a+b)^2$, and $b^2$ is $(a-b)^2$.
$$(a+b)^2-(a-b)^2$$
$$=(a+b+a-b)(a+b-(a-b))$$
$$=(2a)(a+b-a+b)$$
$$=(2a)(2b)$$
$$=4ab$$
A: Since $x^2-y^2=(x+y)(x-y)$, let $x=a+b$ and $y=a-b$. Then,
$$(a+b)^2-(a-b)^2=(a+b+a-b)(a+b-a+b)=(2a)(2b)=4ab$$
A: If you have figured it out then you can ignore this. But I think your confusion is $4ab$ looks like one term, then how is it a factor. Did I get it right?  
If yes then you see, the term $4ab$ is actually $4 * a * b$ where the factors are $4$, $a$ and $b$. So it is the simplest factorized form of your problem.
