document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=290648>Question 290648</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you: Find the other zeroes of f(x)= x^3+5x^2-18x-72 given that one zero is 4?<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 #210305 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=210305\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> divide that equation by (x-4) and you will get a quadratic equation.But with any luck you can factor it. If not the quadratic formula can find the other zeros<BR>\n" );
document.write( "This one can be factored <BR>\n" );
document.write( "(x-4)*(x+3) (x+6) </SPAN>\n" );
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