I have to rationalize a radical again. -please wait for question to be posted-

Question

anonymous · Answer

$$2\sqrt108$$

turingtest · Answer

put 108 into its prime factorization

anonymous · Answer

Which would be 3?

turingtest · Answer

not exactly

anonymous · Answer

:/

turingtest · Answer

prime factorization works like so:
108
since it's even we can divide it by 2 (or "factor out" a 2) so we get
$$108=2\times54$$now since 54 is even we can factor out another 2 to get$$108=2\times2\times27=2^2\times27$$now continue in this way....
what number can we divide 27 by?

anonymous · Answer

Well, 3.

anonymous · Answer

OH WAIT

9

turingtest · Answer

well, you're right both times actually, since 9 is divisible by 3
so anything divisible by 9 is also divisible by 3
make sense?

anonymous · Answer

Yes. I was trying to answer the GCF. :P

turingtest · Answer

sorry I missed a 3 myself :/
let me do that again

turingtest · Answer

$$\sqrt{108}=\sqrt{2^2\times3\times9}=\sqrt{2^2\times3\times3^2}=(2\times3)\sqrt3$$since 27=3^3, and we can only pull out 3^2, we have to leave a 3 inside the radical

turingtest · Answer

so then what is$$2\sqrt{108}$$?

anonymous · Answer

So 6sqrt3? :D

turingtest · Answer

I just showed you the simplification of $$\sqrt{108}$$I think it should be your job to tell me how the 2 in front changes things

turingtest · Answer

hint:$$2\sqrt{108}=2\times(\sqrt{108})$$

anonymous · Answer

That's what I thought I just said. :|

anonymous · Answer

Was $$6\sqrt3$$ wrong? :|

anonymous · Answer

12 root 3 |dw:1337378835775:dw|