document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=693879>Question 693879</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do your simplify: A.&#8730;15v * &#8730;3b<BR>\n" );
document.write( "B. &#8730;5(2&#8730;10 + &#8730;2x)   C. (2-&#8730;6)(1 + &#8730;6)<BR>\n" );
document.write( "Please show steps. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #427631 by </B><A HREF=/tutors/aboutme.mpl?userid=RedemptiveMath><B>RedemptiveMath(80)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RedemptiveMath><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=427631\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1. &#8730;15v * &#8730;3b\r<BR>\n" );
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document.write( "When multiplying radicals, we want to multiply everything underneath the radicals together and keep them under the radical. We multiply whole numbers together and we multiply the radicals together. So, \r<BR>\n" );
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document.write( "&#8730;15v * &#8730;3b<BR>\n" );
document.write( "&#8730;(15v*3b) (multiply everything under radicals)<BR>\n" );
document.write( "&#8730;45bv (keeping everything under the radical).\r<BR>\n" );
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document.write( "Now we need to simplify the number underneath the radical. 45 can be written as a product of 9 * 5, which 9 can be square rooted to get 3 outside of the radical. In other words,\r<BR>\n" );
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document.write( "&#8730;45bv<BR>\n" );
document.write( "&#8730;[(9)(5)(bv)] (9*5*bv = 45bv, so we can write it this way)<BR>\n" );
document.write( "3&#8730;5bv (we can take the square root of any factor of a number that is already underneath a radical as long as it is not a decimal).\r<BR>\n" );
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document.write( "The simplified answer is 3&#8730;5bv. This answer is equivalent to &#8730;45bv and gives the same answer no matter what b and v equal (as long as they remain the same in both expressions). The factor 9 of 45 can be square rooted and brought outside of the radical by multiplication/radical rules. The 5 can't be square rooted into a whole number, so it remains under the radical. B and v remain underneath the radical because they do not have a square sign on them. Variables must have a square sign on them to be brought out from the radical. To check whether or not 3&#8730;5bv is the equivalent form of &#8730;45bv, we can square 3&#8730;5bv and see if equals 45bv (not worrying about the radical). That is,\r<BR>\n" );
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document.write( "(3&#8730;5bv)^2<BR>\n" );
document.write( "(3&#8730;5bv)(3&#8730;5bv) (definition of squaring terms)<BR>\n" );
document.write( "(3*3)(&#8730;5bv*&#8730;5bv) (multiply whole numbers together and radicals together)<BR>\n" );
document.write( "9(&#8730;5bv)^2 (definition of squaring terms)<BR>\n" );
document.write( "9(5bv) (square signs cancel out radicals)<BR>\n" );
document.write( "45bv (multiplication).\r<BR>\n" );
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document.write( "Since the square of 3&#8730;5bv equals 45bv, and the square of &#8730;45bv equals 45bv, 3&#8730;5bv and &#8730;45bv must be equivalent. The first is just the simplified version of the latter. \r<BR>\n" );
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document.write( "2. &#8730;5(2&#8730;10 + &#8730;2x)\r<BR>\n" );
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document.write( "Here we have distributive property to worry about. The rules still stay the same. So,\r<BR>\n" );
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document.write( "&#8730;5(2&#8730;10 + &#8730;2x)<BR>\n" );
document.write( "&#8730;5(2&#8730;10) + &#8730;5(&#8730;2x) (distributive property)<BR>\n" );
document.write( "2(&#8730;5*&#8730;10) + (&#8730;5*&#8730;2x) (multiplying radicals)<BR>\n" );
document.write( "2(&#8730;50) + (&#8730;10x) (multiplying everything under radicals)<BR>\n" );
document.write( "2&#8730;50 + &#8730;10x (remove parentheses won't change expression in this case)<BR>\n" );
document.write( "2&#8730;(25*2) + &#8730;10x (50 can be written as a product of 25 and 2)<BR>\n" );
document.write( "2(5)&#8730;2 + &#8730;10x (the square root of 25 is 5; bring it out and multiply it to any whole number already outside of the radical)<BR>\n" );
document.write( "10&#8730;2 + &#8730;10x (multiply 2 and 5).\r<BR>\n" );
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document.write( "It is dependent on your teacher which way you write it, but the general idea is that the simplified expression is 10&#8730;2 + &#8730;10x. If you were to calculate this expression into a decimal and if you did the same for the original expression &#8730;5(2&#8730;10 + &#8730;2x), the decimal would be the same. To check ourselves to see if 10&#8730;2 is the simplified radical of 2&#8730;50, we do the same thing as we did in the first problem (we must make sure that the square of this term and the square of the original term 2&#8730;50 are the same):\r<BR>\n" );
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document.write( "(10&#8730;2)^2<BR>\n" );
document.write( "(10&#8730;2)(10&#8730;2)<BR>\n" );
document.write( "(10*10)(&#8730;2*&#8730;2)<BR>\n" );
document.write( "100(&#8730;2)^2<BR>\n" );
document.write( "100(2)<BR>\n" );
document.write( "200.\r<BR>\n" );
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document.write( "Is this equal to the square of 2&#8730;50?\r<BR>\n" );
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document.write( "(2&#8730;50)^2<BR>\n" );
document.write( "(2&#8730;50)(2&#8730;50)<BR>\n" );
document.write( "4(&#8730;50)^2<BR>\n" );
document.write( "4(50)<BR>\n" );
document.write( "200.\r<BR>\n" );
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document.write( "It is, so we have simplified 2&#8730;50 correctly.\r<BR>\n" );
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document.write( "3. (2 - &#8730;6)(1 + &#8730;6)\r<BR>\n" );
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document.write( "For this one, we just use FOIL as if this was simple binomial multiplication. So,\r<BR>\n" );
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document.write( "(2 - &#8730;6)(1 + &#8730;6)<BR>\n" );
document.write( "2(1 + &#8730;6) + (-&#8730;6)(1 + &#8730;6) (FOIL)<BR>\n" );
document.write( "2 + 2&#8730;6 + (-&#8730;6) + (-&#8730;36) (when we multiply a whole number and radical together, we just write them side by side like 2&#8730;6).<BR>\n" );
document.write( "2 + 2&#8730;6 - 1&#8730;6 - 1&#8730;36 (we can't forget that there is a -1 in front of &#8730;6 and &#8730;36 when there is a negative sign outside of the radical)<BR>\n" );
document.write( "2 + 1&#8730;6 - &#8730;36 (subtract 2&#8730;6 and 1&#8730;6 because they share the same radical)<BR>\n" );
document.write( "2 + 1&#8730;6 - 6 (the square root of 36 is 6)<BR>\n" );
document.write( "-4 + 1&#8730;6 (combine like terms 2 and -6)<BR>\n" );
document.write( "-4 + &#8730;6 (you don't have to write the 1 in front of &#8730;6 in final answer).\r<BR>\n" );
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document.write( "To check ourselves, we can simply calculate the original expression and the final expression to a decimal and see if they match. If they do, then we've done it correctly (assuming we've simplified as far as we can).  </SPAN>\n" );
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