Question

Okay pleaase look at the comment for the question this box is being dumb

Answer 1

stupid boxes :P

Answer 2

lol even the comment box was being stupid!

Answer 3

factor what is under the radical sign completely
then take out anything you have two of
for instance, what is the prime factorization of 98 ?

Answer 4

\[4\sqrt{2*49}=4*\sqrt{2*7^{2}}=28\sqrt{2}\]

Answer 5

hurry, befor turing erase this....:)

Answer 6

please note that myko is using the method I prescribed, so hopefully you won't have to ask this same type of question again

Answer 7

okay i understand how now but what if there was a number like this.......

Answer 8

is that just an exponent for the 2?

Answer 9

I'm not erasing the above because it does show steps
if @ParisWinter does not understand what myko did, they should ask for explanation.
In any case, I hope not to see the exact same type of question again... oh wait, here we go again

what is the prime factorization of 128 ?

Answer 10

64 i understand that but i dont understand what to do with the 3

Answer 11

64 is not the prime factorization of 128
do you know how to factor?

Answer 12

, ill figure it out but what do i need to do with 3

Answer 13

the 3 just tells you what you can take out; ^3

Answer 14

\[\sqrt[n]{a^n}=a\]

Answer 15

take out of 128?

Answer 16

take out from under the radical; does 128 have any perfect ^3s in it?

Answer 17

no

Answer 18

I think it does
you should factor it and see

Answer 19

128 = 2 * 64
= 2*2*32
= 2*2*2*16
=2*2*2*2*8
=2*2*2*2*2*4
=2*2*2*2*2*2*2
=(2*2*2)*(2*2*2)*2

Answer 20

okay i dont see a three anywhere in that.

Answer 21

do you see any groups that contain 3 of the same numbers?

Answer 22

\[a*a*a = a^3\]