document.write( "Question 621297: can you reduce 48 sqrt (3)
\n" ); document.write( "and can someone please give me a few helpful tips on remembering how to reduce them because i have alot of trouble.
\n" ); document.write( "thanks!:)
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #390703 by fcabanski(1391) $\"\"$ $\"About$

You can put this solution on YOUR website!
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.

\n" ); document.write( " $\"sqrt%2816%29\"$ 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.

\n" ); document.write( "Since there are two pairs of 2's pull two 2's out of the square root: 2*2* $\"sqrt%280%29\"$ = 4.

\n" ); document.write( " $\"48sqrt%283%29\"$ cannot be factored because 3 is already a prime number.

\n" ); document.write( " $\"sqrt%288%29\"$ | 8 factors to 2*2*2. Pull out two of the 2's, so one remains inside the square root:

\n" ); document.write( " $\"2sqrt%282%29\"$
\n" ); document.write( "

Hope the solution helped. Sometimes you need more than a solution. Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.) \n" ); document.write( "

\n" );