document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=168539>Question 168539</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 3x^3+x^2-3x-1=0   (The ^3 and ^2 and exponents) \r<BR>\n" );
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document.write( "I have tried to factor and pull out the x but I am getting confused with the two different power. I got x=-1, x=-1/3 and x=1 but i dont think i need three answers </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #124211 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=124211\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 3x^3+x^2-3x-1=0 <BR>\n" );
document.write( "x^2(x+1) -3(x+1) = 0<BR>\n" );
document.write( "(x+1)(x^2-3) = 0<BR>\n" );
document.write( "x = -1 or x=sqrt(3) or x=-sqrt(3)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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