Question 972180
<font face="Times New Roman" size="+2">


According to the rational roots theorem, the possible rational zeros of your function are *[tex \Large \pm 1] and *[tex \Large \pm 2].  If a polynomial function with integer coefficients has a rational root, then that root will be of the form *[tex \Large \frac{p}{q}] where *[tex \Large p] divides *[tex \Large a_n], the constant term, and *[tex \Large q] divides *[tex \Large a_0] the lead coefficient.


Synthetic division with each of *[tex \Large 1], *[tex \Large -1], *[tex \Large 2], and *[tex \Large -2] yields a remainder.  Hence, this function has no rational zeros.


The graph of the function:


{{{drawing(
500, 500, -5,5,-5,5,
grid(1),
graph(
500, 500, -5,5,-5,5,
x^3+2x^2+x-2))}}}


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.


By the way, the derivative does you no good because the local extrema are nowhere near (relatively speaking) the *[tex \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 href="http://en.wikipedia.org/wiki/Newton%27s_method">Wikipedia Newton's Method</a>


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 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.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \