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.

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