document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1023110>Question 1023110</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Factor. Began by factoring out the common monomial\r<BR>\n" );
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document.write( "12x^2 - 68x + 80\r<BR>\n" );
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document.write( "NOTE:<BR>\n" );
document.write( "      ^2 means x squared </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #638632 by </B><A HREF=/tutors/aboutme.mpl?userid=god2012><B>god2012(113)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=god2012><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=god2012><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=638632\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 12x^2 - 68x + 80 \r<BR>\n" );
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document.write( "Here the factors are derived as below:<BR>\n" );
document.write( "1. Try to find 2 numbers whose sum is equivalent to the middle number 68 and product is the product of the first and the last constants , in this case they are 12 and 80.\r<BR>\n" );
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document.write( "Sum = -68 = (-48) + (-20)<BR>\n" );
document.write( "Product = 12*80 = 960 = 48*20 = (-48) * (-20)\r<BR>\n" );
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document.write( "12x^2 - 68x + 80 = 12x^2 - 48x - 20x + 80<BR>\n" );
document.write( " = 12x(x - 4) - 20(x - 4)<BR>\n" );
document.write( " = (x - 4)(12x - 20)<BR>\n" );
document.write( " = (x - 4)4(3x - 5)\r<BR>\n" );
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document.write( "12x^2 - 68x + 80 = 4(x - 4)*(3x - 5) </SPAN>\n" );
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