Question 621297
When a square root is involved factor whatever is inside the square root sign to prime numbers.  Any prime number repeated twice, factor out (once).  If it repeats twice it is a perfect square.<P>
{{{sqrt(16)}}} for example.  Factor to 2*8.  8 can be further factored to 2*4, so now we have 2*2*4.  4 can be factored to 2*2 so now we have 2*2*2*2.<P>
Since there are two pairs of 2's pull two 2's out of the square root:  2*2* {{{sqrt(0)}}} = 4.<P>
{{{48sqrt(3)}}} cannot be factored because 3 is already a prime number.<P>
{{{sqrt(8)}}} | 8 factors to 2*2*2.  Pull out two of the 2's, so one remains inside the square root:<P>
{{{2sqrt(2)}}}
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