document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=972180>Question 972180</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How would you go about finding the zeroes for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29=x%5E3%2B2x%5E2%2Bx-2 ALIGN=MIDDLE  ALT=\"f%28x%29=x%5E3%2B2x%5E2%2Bx-2\"   DISABLED_style= \"max-width:90%; height:auto;\" >? I've tried to solve for the zeroes using synthetic division with positive and negative 2 and 1 but all of my answers have had remainders.\r<BR>\n" );
document.write( "\n" );
document.write( "I've tried using the derivative <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B4x%2Bx ALIGN=MIDDLE  ALT=\"3x%5E2%2B4x%2Bx\"   DISABLED_style= \"max-width:90%; height:auto;\" > to solve for the zeroes using negative 1 and 1/3, but I know that's not correct. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #594555 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=594555\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "According to the rational roots theorem, the possible rational zeros of your function are <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pm 1\"> and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pm 2\">.  If a polynomial function with integer coefficients has a rational root, then that root will be of the form <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{p}{q}\"> where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large p\"> divides <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large a_n\">, the constant term, and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large q\"> divides <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large a_0\"> the lead coefficient.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Synthetic division with each of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 1\">, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large -1\">, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 2\">, and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large -2\"> yields a remainder.  Hence, this function has no rational zeros.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The graph of the function:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28%0D%0A500%2C+500%2C+-5%2C5%2C-5%2C5%2C%0D%0Agrid%281%29%2C%0D%0Agraph%28%0D%0A500%2C+500%2C+-5%2C5%2C-5%2C5%2C%0D%0Ax%5E3%2B2x%5E2%2Bx-2%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "reveals that there is only one real root and therefore 2 complex roots.  Since there are no rational roots, the real root must be irrational.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "By the way, the derivative does you no good because the local extrema are nowhere near (relatively speaking) the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\">-axis, and you have nothing that tells you the value of the derivative at the intercept.  You can use the first derivative in Newton-Raphson to approximate the real zero as precisely as you like.  <a  rel=nofollow HREF=\"http://en.wikipedia.org/wiki/Newton%27s_method\">Wikipedia Newton's Method</a>\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "In order to find the exact value of the roots you need to use the formula for the solution of the general cubic.  If you are interested, look here: <a  rel=nofollow HREF=\"http://en.wikipedia.org/wiki/Cubic_function\">Wikipedia Cubic Function</a>.  You can root through this mess to find the solution yourself if you are that big a glutton for punishment.  If your professor/teacher/instructor insists that you do this, I would drop the course and go learn from someone who is not a direct descendant of the Marquis de Sade.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \ \\r\"> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );