2
$\begingroup$

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

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 '14 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 '14 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 '14 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 '14 at 17:25
  • 1
    $\begingroup$ I am sorry. It was not of my business. $\endgroup$ – Jika Jun 14 '14 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 '14 at 17:31
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.