document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=697326>Question 697326</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find all real zeros of the function: g(x)=(x^3)-2x+1 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #430044 by </B><A HREF=/tutors/aboutme.mpl?userid=Positive_EV><B>Positive_EV(69)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Positive_EV><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=Positive_EV><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=430044\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A good place to start looking for solutions to cubic functions is to use the Rational Root theorem to try to find rational roots, then use any root found to find a depressed quadratic equation. Any rational roots of a polynomial must be in the form +/- p/q, where p is any number that divides evenly into the constant term, and q is any number that divides evenly into the coefficient of the term with the highest degree. In this case, the constant term is 1 and the coefficient of the x^3 term is also 1, so the only possible rational roots are 1 and -1. Plugging these into the expression yields:<br>\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%5E3+-+2%281%29+%2B+1+=+0 ALIGN=MIDDLE  ALT=\"1%5E3+-+2%281%29+%2B+1+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so 1 is a zero.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%29%5E3+-+2%28-1%29+%2B+1+=+2 ALIGN=MIDDLE  ALT=\"%28-1%29%5E3+-+2%28-1%29+%2B+1+=+2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so -1 is not a zero.<br>\r<BR>\n" );
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document.write( "Since one zero is now known, you can use synthetic division to find a depressed quadratic equation:<br>\r<BR>\n" );
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document.write( "<TD >1</TD>\r\n" );
document.write( "<TD ></TD>\r\n" );
document.write( "<TD >1</TD>\r\n" );
document.write( "<TD >0</TD>\r\n" );
document.write( "<TD >-2</TD>\r\n" );
document.write( "<TD >1</TD>\r\n" );
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document.write( "<TD ></TD>\r\n" );
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document.write( "<TD >1</TD>\r\n" );
document.write( "<TD >1</TD>\r\n" );
document.write( "<TD >-1</TD>\r\n" );
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document.write( "<TD ></TD>\r\n" );
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document.write( "<TD >1</TD>\r\n" );
document.write( "<TD >1</TD>\r\n" );
document.write( "<TD >-1</TD>\r\n" );
document.write( "<TD >0</TD>\r\n" );
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document.write( "From this, looking at the bottom row of the synthetic division, the depressed equation is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2+%2B+x+-+1+=+0 ALIGN=MIDDLE  ALT=\"x%5E2+%2B+x+-+1+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Use the quadratic formula to solve for x, where a = 1, b = 1, c = -1:<br>\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+ ALIGN=MIDDLE  ALT=\"x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-1+%2B-+sqrt%285%29%29%2F2 ALIGN=MIDDLE  ALT=\"x+=+%28-1+%2B-+sqrt%285%29%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br>\r<BR>\n" );
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document.write( "So the three zeros are 1 and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1+%2B-+sqrt%285%29%29%2F2 ALIGN=MIDDLE  ALT=\"%28-1+%2B-+sqrt%285%29%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >. </SPAN>\n" );
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