document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=203973>Question 203973</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Ok, so I was told to 'factor the trinomial completly' with this problem...\r<BR>\n" );
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document.write( "x^7-x^6-64x+64 .....\r<BR>\n" );
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document.write( "I think I am sapose to combine like terms so it would be in the form x^2+bx+c where x is one. <BR>\n" );
document.write( "So I combined them and got x-64x+64 and I know I have to factor but I am not sure if I can....because the product is a*c=64 and the sum is -64 and I am not sure what two numbers multiply together that equal 64 that also add together to equal -64. <BR>\n" );
document.write( "help? \r<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 #153905 by </B><A HREF=/tutors/aboutme.mpl?userid=nerdybill><B>nerdybill(7384)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nerdybill><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=nerdybill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=153905\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To factor, use \"grouping\" technique:<BR>\n" );
document.write( "x^7-x^6-64x+64<BR>\n" );
document.write( "= (x^7-x^6)-(64x-64)<BR>\n" );
document.write( "= x^6(x-1)-64(x-1)<BR>\n" );
document.write( "= (x^6-64)(x-1)<BR>\n" );
document.write( "But wait, we still factor left term:<BR>\n" );
document.write( "= (x^3-8)(x^3+8)(x-1)<BR>\n" );
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