Question 834605
3 complex roots, due to the fundamental theorem of algebra.

The cubic must have at least one real root. It cannot have exactly two real roots, as the third would have non-zero imaginary part, which doesn't work (since the sum of the roots is real). So the cubic has either 1 or 3 real roots (in actuality, one).

Possible rational roots are of the form *[tex \frac{a}{b}] where a divides 6 and b divides 3.