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A thingy

I need help with solving this one.. How do I factor the a^4-b^4?
$\frac{a^4-b^4}{4a^2+4b^2} : (a+b)$

The difference of two squares:

$a^{4}-b^{4}=(a^{2})^{2}-(b^{2})^{2}=(a^{2}+b^{2})(a^{2}-b^{2})$

what if x=a^2 and y=b^2.
then you'd have x^2-y^2.
do you know how to factor that one?
then you can just replace the values for x and y back in and you've got your answer.

Quote:

Originally Posted by asterisk_man

what if x=a^2 and y=b^2.
then you'd have x^2-y^2.
do you know how to factor that one?
then you can just replace the values for x and y back in and you've got your answer.

So it's (x-y)(x+y) and after replacing $(a^2-b^2)(a^2+b^2)$ ?
Thanks a lot, appreciate it!

you got it. Then I'm sure you saw that you can then cancel the (a^2+b^2) with the (a^2+b^2) on the bottom to get (a^2-b^2)/4

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