SOLUTION: How do i go about factoring: sqrt(128) I looked online on how to solve radicals and i found a simple tutorial, i looked for questions and found the one above, what i did was: s

Question 770818: How do i go about factoring: sqrt(128)
I looked online on how to solve radicals and i found a simple tutorial, i looked for questions and found the one above, what i did was:
sqrt (128)
sqrt (128)= sqrt(64 x 2)
sqrt (128)= sqrt(64) x sqrt(2)=sqrt(16 x 4) sqrt (2)
sqrt (128)= sqrt(64) x sqrt(2)=sqrt(16) x sqrt (4) x sqrt (2)
~sqrt(16) x sqrt(4) x sqrt(2)-> sqrt(4x4) x sqrt(4x1 or 2x2?) x sqrt(2)
Some questions i have:

1: Would i divide the "sqrt (4) by 4 because it is a perfect square to make it "sqrt (4 x 1), or by 2 because 2x2 is 4? and sqrt(4) is 2?
2: I get confused as to what it means as to multiplying the numbers that stay out to the outside number, if you understand this question, could you explain it?
3: Is there a simpler way to solve this? i know of a tree/branching format but it looks confusing.
4: Which method sets me up better to factor, Box or F.O.I.L, why? My teacher had said F.O.I.L but i am not sure as to why. I prefer the Box method because it seems to involve less steps and gets me to an answer right away.
Any feedback is appreciated.
Answer by oscargut(2103) (Show Source): You can put this solution on YOUR website!
sqrt (128)= sqrt(64 x 2) =sqrt(8^2 x 2) = sqrt(8^2) sqrt)2) = 8sqrt(2)
Answer:8sqrt(2)
You can ask me more at: mthman@gmail.com
Thanks
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