Calculus
posted by Untamed .
Simplify. State the nonpermissible values.
4a^21/4a^216*2a/2a1
This is what I did to try and solve it out:
(2a+1)(2a1)/(2a+4)(2a4)*2a/(2a1)
I cancelled out the (2a1)'s and don't get how to solve.
I'm supposed to get the answer :
2a1/4(a+2)
But I don't get how to get that^.

an (2a4)=2(a2) or 2(2a) which will cancel out your 2a,
then do the same for the 2a+4, and this should give you your answer 
Thanks so much =)

I read that as
(4a^21)/(4a^216) * (2a)/(2a1) , my brackets are necessary
= (2a+1)(2a1)/( 4(a+2)(a2) * (2a)/(2a1)
the (2a)/(a2) are opposite, they will give you the 1
so..
= (2a+1)/(4(a+2)) , again you have to use brackets to write it on here. 
Thanks Reiny =) But I to get everything clear, the 2a and the a2 cancel right? After I cancel those I am left with (2a+1) as the numerator:S Sorry I am just confused. Also did you factor out the 2a+4 and 2a4 to get 4(a+2)(a2)? Could you please explain step by step?