document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=479551>Question 479551</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I'm having trouble factoring out this problem, I didn't know where to post a factoring question so I apologize if this is the wrong place, but im trying to factor out x^2 - 15x - 2 I tried isolating the coefficient and constant and cross multiply to get the two numbers that = -15x, but I'm stumped lol, I'd appreciate a hand! Thannks! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #328563 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=328563\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x^2 - 15x - 2\r<BR>\n" );
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document.write( "This can be done in many ways\r<BR>\n" );
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document.write( "x^2-15x+(7.5)^2-2=0\r<BR>\n" );
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document.write( "x^2-15x+(7.5)^2=2+(7.5)^2\r<BR>\n" );
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document.write( "(x-7.5)^2= 2+56.25<BR>\n" );
document.write( "(x-7.5)^2=58.25\r<BR>\n" );
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document.write( "Take the square root of both sides\r<BR>\n" );
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document.write( "(x-7.5)= +/- sqrt(58.25)<BR>\n" );
document.write( "(x-7.5)= 7.63\r<BR>\n" );
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document.write( "x= 7.5+7.63= 15.13\r<BR>\n" );
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document.write( "OR<BR>\n" );
document.write( "x= 7.5-7.63= -0.13\r<BR>\n" );
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document.write( "m.ananth@hotmail.ca </SPAN>\n" );
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