I can tell you how to factorise a Cubic polynomial. This would be a long lecture, so after reading this you try out with some polynomials.
Let's Start:
A third degree Polynomial is in the form of
$$x^3 + bx^2+cx+d$$
Let the roots be $\alpha,\beta,\gamma$
Do you Know the symmetric notation:
$$ x^3 + (\sum_{}^{} \alpha )x^2 + (\sum_{}^{} \alpha\beta )x + \alpha\beta\gamma $$
Here,
$$\sum_{}^{} \alpha = \alpha+\beta+\gamma$$
$$\sum_{}^{} \alpha\beta = \alpha\beta + \beta\gamma +\gamma \alpha$$
$$\alpha\beta\gamma$$
What I am doing here is I am just expressing the coefficients in terms of roots just as we do in Quadratic Equation: $x^2+ (\alpha\beta)x+\alpha\beta$.
Or You can write ie like this also:
$$ x^3 + (\alpha+\beta+\gamma)x^2 + (\alpha\beta + \beta\gamma +\gamma \alpha )x + \alpha\beta\gamma $$.
What can we get from these expressions??
We see that our $d$ is product of all roots, $b$ is Sum of all roots.
So, you can take these coefficients and try to guess the values.
I shall illustate this. In your case:
$$x^3 - 3x - 2$$
$$x^3 +0x^2- 3x - 2$$.
Here,
$$\alpha\beta\gamma = -2$$
Factors of -2 are $\pm1,\pm2$
Try to Guess here:
$$(-2)\cdot(+1)\cdot(+1)=-1$$
$$(-2)+(+1)+(+1)=0=\alpha+\beta+\gamma$$
Try for:
$$\alpha\beta + \beta\gamma +\gamma \alpha= (-2)(+1)+(+1)(+1)+(+1)(-2)=-2 + 1 +(-2) = -3 $$
Then You got all your zeroes:
Now Put them back :
$$(x-2)(x+1)(x+1)$$
Hence You got it.
Note: We are not changing any signs here because we have not changed any signs in the symmetric notation.
We give some work to "intutition" here!!
If there is existing $a$ here, then you divide that :
$$x^3+\frac{b}{a} x^2+\frac{c}{a} x + \frac{d}{a}$$
Or I say you use THE CUBIC FORMULA.
http://mathworld.wolfram.com/CubicFormula.html
http://www.math.vanderbilt.edu/~schectex/courses/cubic/
These seem scary. There are some books which say how to solve using cubic formula.
I use the above method.
Hope this helps!!