document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=39215>Question 39215</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (2a+4)(4a^2-3a+4)= </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #24675 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=24675\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For working out a problem like<BR>\n" );
document.write( "(2a+4)(4a^2-3a+4),<BR>\n" );
document.write( "we multiply every term in the first factor by every term in the second, giving us six terms, which we will combine afterwards...so<BR>\n" );
document.write( "(2a+4)(4a^2-3a+4) =<BR>\n" );
document.write( "8a^3 - 6a^2 + 8a + 16a^2 - 12a + 16 =<BR>\n" );
document.write( "8a^3 + 10a^2 - 4a + 16 </SPAN>\n" );
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