Question

what are the roots of x^3-2x^2-2x-3=0

Answer 1

Well, first of all there is a formula for solving polynomials of third degree, but no one in high school is expected to know that.
That means that there is usually a 'trick' that allows factoring to first or second degrees polynomials or shaping it to form a second degree polynomial.

Anyway, say we have:
$$
x^3 - 2x^2 - 2x -3 = 0 \\
$$
Well, first you can say:
$$
x = 3 \\
3^3 - 2 \cdot 3^2 - 2 \cdot 3 - 3 = 0 \\
3 \cdot 9 - 2 \cdot 9 - 6 - 3 = 0 \\
9 - 9 = 0
$$So \(x = 3\) is one solution. Now we can say:
$$
x = 3 \implies x - 3 = 0 \\
$$And try to factor (x-3) out of the expression:
$$
x^3 - 2x^2 - 2x -3 = 0 \\
x^2(x - 3) + x^2 - 2x - 3 = 0 \\
x^2(x - 3) + x(x - 3) + (x - 3) = 0 \\
(x-3)(x^2 + x + 1) = 0
$$ The other solutions will be given by the second part, so:
$$
x^2 + x + 1 = 0 \\
x_{1,2} = \frac{-1 \pm \sqrt{(-1)^2 - 4}}{2} = \frac{-1 \pm \sqrt{-3} }{2}
$$And since the we have a square root of a negative number that means that the other two solutions are complex numbers.
If you're supposed to work with real numbers then the only solution is x=3.
For reference:
http://www.wolframalpha.com/input/?i=x%5E3-2x%5E2-2x-3%3D0

Here is a tutorial showing 2 ways facing such problems:
http://www.wikihow.com/Factor-a-Cubic-Polynomial
The second is what used here