document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=119560>Question 119560</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello out there.\r<BR>\n" );
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document.write( "Can someone help me and my son solve this equation?  They are asking for the \"real solutions\" to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E3-13x=12 ALIGN=MIDDLE  ALT=\"x%5E3-13x=12\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Please and much thanks.  It's been a long time since I had to remember this stuff :-) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #87609 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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/cgi-bin/show-face.mpl?user=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=87609\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> subtracting 12 gives x^3-13x-12=0 __ substituting -1 for x satisfies the equation, so x+1 is a factor\r<BR>\n" );
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document.write( "dividing x^3-13x-12 by x+1 gives x^2-x-12, which factors into (x-4) and (x+3)\r<BR>\n" );
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document.write( "so x^3-13x-12=0 factors into (x+1)(x+3)(x-4)=0\r<BR>\n" );
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document.write( "setting each of the factors equal to zero, gives the \"real\" values of x __ -1, -3, 4 </SPAN>\n" );
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