document.write( "Question 425297: simplify the radical exspression \"5sqrt%2824%29+multiply+by+3sqrt%2810%29\"
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Algebra.Com's Answer #296235 by jsmallt9(3758)\"\" \"About 
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\"5sqrt%2824%29%2A3sqrt%2810%29\"
\n" ); document.write( "When multiplying radicals you can either
  • Simplify the radicals, multiply and then simplify the multiplied radical; or
  • Multiply and then simplify the radical.

\n" ); document.write( "Even though the first approach has two simplifications instead of just une, it can actually be easier this way when the expressions are complex. Your expression is not very complex so we will use the second approach.

\n" ); document.write( "Sine the expression is all multiplications, we can use the Commutative and Associative Properties to rearrange the order and grouping:
\n" ); document.write( "\"%285%2A3%29%28sqrt%2824%29%2Asqrt%2810%29%29\"
\n" ); document.write( "This first group, 5*3, is 3rd grade multiplication. To multiply the square roots we use a property of radicals, \"root%28a%2C+p%29%2Aroot%28a%2C+q%29+=+root%28a%2C+p%2Aq%29\", to multiply. SO we get:
\n" ); document.write( "\"15%2Asqrt%2824%2A10%29\"
\n" ); document.write( "or
\n" ); document.write( "\"15%2Asqrt%28240%29\"

\n" ); document.write( "Now we simplify the square roots we look for perfect square factors of the radicand. (The expression inside a radical is called a radicand.) 240 definitely has one (or more) perfect square factors:
\n" ); document.write( "\"15%2Asqrt%284%2A60%29\"
\n" ); document.write( "60 also has a prefect square factor:
\n" ); document.write( "\"15%2Asqrt%284%2A4%2A15%29\"
\n" ); document.write( "Now we use the same property of radicals we used earlier, only in reverse, to split up the square root:
\n" ); document.write( "\"15%2Asqrt%284%29%2Asqrt%284%29%2A15%29\"
\n" ); document.write( "The square roots of the perfect squares simplify:
\n" ); document.write( "\"15%2A2%2A2%2A15%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"60sqrt%2815%29\"
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