3
$\begingroup$

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?

$\endgroup$
4
  • 5
    $\begingroup$ Simplify the factors. $\endgroup$
    – user63181
    Jun 14, 2014 at 17:18
  • $\begingroup$ and that is 4ab. $\endgroup$
    – Edwin_R
    Jun 14, 2014 at 17:19
  • $\begingroup$ you simply open the brackets and simplify. $\endgroup$
    – SA-255525
    Jun 14, 2014 at 17:22
  • 1
    $\begingroup$ What you did is fine. But you can do more. Follow user63181's advice (for instance, $[(a+b)-(a-b)]=2b$). $\endgroup$ Jun 14, 2014 at 17:36

5 Answers 5

6
$\begingroup$

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 $(a+b)^2 - (a-b)^2 = [(a+b)-(a-b)][(a+b)+(a-b)]$

$\hspace{2cm} = [a+b-a+b][a+b+a-b]$

$\hspace{2cm} = [a-a+b+b][a+a+b-b]$

$\hspace{2cm} = [2b][2a]$

$\hspace{2cm} = 4ab$

$\endgroup$
2
  • $\begingroup$ But the question asks for the factors. How is this factoring? $\endgroup$
    – Cetshwayo
    Jun 14, 2014 at 17:26
  • $\begingroup$ @Utvecklaochförenkla, Now you see, if you are getting the result. You have done everything. You just need to carry out one more step. $\endgroup$
    – L.K.
    Jun 14, 2014 at 17:34
4
$\begingroup$

$$(a + b)^2 - (a-b)^2 = (a+b)(a+b) - (a+b)(a-b) $$ $$= (a+b)((a+b) - (a - b)) $$ $$ = (a + b)(a + b - a + b) $$ $$= (a+b)(2b)$$

POST EDIT: $$(a+b)^2 - (a-b)^2 = a^2 +2ab + b^2 -(a^2 - 2ab + b^2) = 2ab - (-2ab) = 4ab$$

$\endgroup$
4
  • 1
    $\begingroup$ Sir, it is $(a-b)^2$ not $a^2-b^2$. $\endgroup$
    – Jika
    Jun 14, 2014 at 17:23
  • 2
    $\begingroup$ @Jika I'm not a "sir" and I answered prior to the OP's edit. (And after the edit. $\endgroup$
    – amWhy
    Jun 14, 2014 at 17:25
  • 1
    $\begingroup$ I am sorry. It was not of my business. $\endgroup$
    – Jika
    Jun 14, 2014 at 17:29
  • 1
    $\begingroup$ Do not fret, @Jika. If I weren't on the page when you commented, your comment would have "pinged me" (the system would notify me), in case I wanted to update. So don't second guess your comment. $\endgroup$
    – amWhy
    Jun 14, 2014 at 17:31
4
$\begingroup$

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 $(a-b)^2$. $$(a+b)^2-(a-b)^2$$ $$=(a+b+a-b)(a+b-(a-b))$$ $$=(2a)(a+b-a+b)$$ $$=(2a)(2b)$$ $$=4ab$$

$\endgroup$
3
$\begingroup$

Since $x^2-y^2=(x+y)(x-y)$, let $x=a+b$ and $y=a-b$. Then, $$(a+b)^2-(a-b)^2=(a+b+a-b)(a+b-a+b)=(2a)(2b)=4ab$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .