Can a cubic that crosses the x axis at three points have imaginary roots? I have a cubic polynomial, $x^3-12x+2$ and when I try to find it's roots by hand, I get two complex roots and one real one. Same, if I use Mathematica. But, when I plot the graph, it crosses the x-axis at three points, so if a cubic crosses the x-axis a three points, can it have imaginary roots, I think not, but I might be wrong.

You may have gotten the answer $\sqrt[3]{(-3i\sqrt{7}-1)}+ \sqrt[3]{(3i\sqrt{7}-1)}$ when you computed it by hand. The three real solutions can be presented as follows, rounded as it appears that this cubic equation may not be able to be solved with radicals
$$(-1.772303408...+i0.9267905...)+(-1.772303408...-i0.9267905...)=-3.544...$$
$$(0.083527587...+i1.998255024...)+(0.083527587...-i1.998255024...)=.1670...$$
$$(1.688775821...+i1.071464524...)+(1.688775821...-i1.071464524...)=3.377...$$
Each of the above numbers in parentheses is a cube root of 
 $$\sqrt[3]{(-3i\sqrt{7}-1)}$$ and $$\sqrt[3]{(3i\sqrt{7}-1)}$$  respectively. You can see that each $Im(z)$ cancel to arrive at the real number solution. I believe this is called casus irreducibilis which you can check out on Wikipedia  http://en.wikipedia.org/wiki/Casus_irreducibilis. You can see one method of computing the cube roots of these nested cubic radicals on Stack Exchange Question 16331.
