document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=67430>Question 67430</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (x^2-2x-15)(x-4)=0 solve by finding roots<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #47961 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=47961\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (x^2-2x-15)(x-4)=0 solve by finding roots\r<BR>\n" );
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document.write( "Note that x^2-2x-15 has factors of (x-5)(x+3). How did I know this??   I determined that it has factors by inspection and by knowing that when a quadratic (Ax^2+Bx+C=0) has the A coefficient equal to 1, then the B coefficient is equal to the sum of the factors of the C coefficient, if it is factorable and in this case (-5)(+3) works.\r<BR>\n" );
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document.write( "So we have:\r<BR>\n" );
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document.write( "(x-5)(x+3)(x-4)=0\r<BR>\n" );
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document.write( "x=5<BR>\n" );
document.write( "x=-3<BR>\n" );
document.write( "x=4\r<BR>\n" );
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document.write( "Hope this helps-------ptaylor\r<BR>\n" );
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