document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=93562>Question 93562</A>: <I><SPAN STYLE=\"word-break: break-all;\"> factor completely\r<BR>\n" );
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document.write( "16x^2-2x-3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #68140 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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=68140\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> DEFINE ALL FACTORS OF 16=1,-1,2,-2,4,-4,8,-8,16&-16<BR>\n" );
document.write( "DEFINE ALL FACTORS OF -3=1,-1,3&-3<BR>\n" );
document.write( "NOW FIND ONE SET OF FACTORS THAT WHEN MULTIPLIED AND THEN ADDED=(-2).<BR>\n" );
document.write( "PICK ONE FACTOR FROM 16 & ONE FROM -3 MULTIPLY THESE FACTORS. NOW MULTIPLY THE COMPANION FACTOR OF 16 & -3. MULTIPLY THESE FACTORS. NOW ADD THESE TWO PRODUCTS. THEY SHOULD EQUAL -2.<BR>\n" );
document.write( "THERE IS ONLY ONE SET OF FACTORS THAT SATISFY THIS PROBLEM.<BR>\n" );
document.write( "8*-1=-8 & 2*3=6 [8*2=16 & -1*3=3] THEN -8+6=-2 WHICH IS THE MIDDLE TERM.<BR>\n" );
document.write( "6x^2-2x-3<BR>\n" );
document.write( "(8x+3)(2x-1)<BR>\n" );
document.write( "THIS LOOKS COMPLICATED BUT WITH A LITTLE PRACTICE YOU CAN MASTER IT. \"DON'T GIVE UP THE SHIP JUST BECAUSE THERE IS A WHITE CAP ON THE HORIZEN\". </SPAN>\n" );
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