document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=175687>Question 175687</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Factor the Polynomials:\r<BR>\n" );
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document.write( "x^3+y^3-x^2y-xy^2\r<BR>\n" );
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document.write( "81x^4-16y^4\r<BR>\n" );
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document.write( "x^3+2x^2-255x\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 #130936 by </B><A HREF=/tutors/aboutme.mpl?userid=srcedwards><B>srcedwards(3)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=srcedwards><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=130936\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1)  Factor by grouping:\r<BR>\n" );
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document.write( "x^3 + y^3 - x^2y - xy^2 =<BR>\n" );
document.write( "(x^3 + y^3) + -(x^2y + xy^2).  The first is a sum of cubes:<BR>\n" );
document.write( "[(x + y)(x^2 - xy + y^2)] + [-xy(x + y)].  They both have an x + y in common<BR>\n" );
document.write( "(x + y)[(x^2 - xy + y^2) + (-xy)].  Now simplify:<BR>\n" );
document.write( "(x + y ( x^2 - 2xy + y^2)\r<BR>\n" );
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document.write( "2)  This is a difference of squares:\r<BR>\n" );
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document.write( "81x^4 - 16y^4 =<BR>\n" );
document.write( "(9x^2)^2 - (4y^2)^2 = (9x^2 - 4y^2)(9x^2 + 4y^2).  But the first is a difference of squares, so that can be factored again:\r<BR>\n" );
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document.write( "(3x - 2y)(3x + 2y)(9x^2 + 4y^2)\r<BR>\n" );
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document.write( "3)  Factor out a common factor, x:\r<BR>\n" );
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document.write( "x^3 + 2x^2 -255x =<BR>\n" );
document.write( "x(x^2 + 2x - 255).  Can this be factored again?  255 = 15 * 17 which has a difference of two (the middle coefficient), so it can be factored:<BR>\n" );
document.write( "x(x - 15)(x - 17) </SPAN>\n" );
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