How to factorize $x^3-3x-2$ with a technique similar to the cross factoring method I usually factor polynomials using the criss cross method. However, I noticed that in the derivatives section that I am currently studying that there are quite a few problems whch are cubic polynomials, such as $$x^3-3x-2$$ I know that the answer is $$(x+1)^2(x-2)$$ but how do I actually do the work to arrive at that answer. 
I need a technique that works. Please do not comment if you do not understand that I am a pre-calculus student. Thus, the less math jargon the better. Cheers!
 A: If you add and subtract $x^2$ in your polynomial, it will be easy to then factorise it using middle term splitting:$$x^3+x^2-x^2-3x-2$$ $$x^2(x+1)-(x^2+3x+2)$$$$x^2(x+1)-(x+1)(x+2)$$$$(x+1)(x^2-x-2)$$ $$(x+1)^2(x-2)$$
But this answer is useful only in this specific case,  and not all cubics can be factorised so easily, in fact most of those can't even be factorised without knowing the roots.
A: Factoring these types of polynomials can be a little tricky, and it uses a lot of results. I will outline the method to give you a push in the right direction.
In order to factor this, you need to know what it means to be a zero of a function. We say that $c$ is a zero of $f(x)$ if $f(c) = 0$, in other words, you plug in $c$ and get zero. Another equivalent definition of being a zero is: $c$ is a zero if $(x - c)$ is a factor of $f(x)$. This can help you factor polynomials.
Next, you need the Rational Root Test. The rational root test says if you want to find the zeros of a polynomial $f(x) = Ax^n + \cdots + Bx + C$, the possible rational zeros are
$$ x = \frac{p}{q},$$
where $p$ is a factor of $C$ (your constant term) and $q$ is a factor of $A$ (the leading coefficient).
Now to the method: First, you find the possible rational zeros of $f(x) = x^3 - 3x - 2$ using the rational root test. This will give you:
$$ x = \frac{\mbox{factors of -2}}{\mbox{factors of 1}} = \pm1, \pm2.$$
Now, you check which one of these four numbers (it can be more than one of them) is an actual zero by using the first definition: plug each one into $f(x)$ and see if you get zero. It turns out that $x = 2$ will work since
$$f(2) = 2^3 - 3(2) - 2 = 8 - 6 - 2 = 0.$$
After that, you use the second definition of what it means to be a zero. Since $2$ is a zero, this means $(x-2)$ is a factor of $f(x)$. So, we have
$$ f(x) = (x-2)g(x),$$
where $g(x)$ is some other polynomial we have to find.
To find $g(x)$, you divide both sides by $(x-2)$
$$g(x) = \frac{f(x)}{(x-2)}.$$
So... to finish this problem, you have to use polynomial division and divide $x^3 - 3x - 2$ by $x-2$. The result of this division will give you the $g(x)$ you need to factor your polynomial. You can either use long division or synthetic division to figure it out. 
A: First of all, if $f(x)=x^3-3x-2=0$ has a rational root, then by the rational root theorem it has to be either 
$$\pm 1,\ \ \pm 2.$$
Then, we can see
$$f(-1)=f(2)=0.$$
Hence, by the factor theorem, we know that $f(x)$ has factors $(x+1)$ and $(x-2)$. So, you can divide $f(x)$ by $(x+1)(x-2)$ to get $(x+1)$. 
Thus, we have
$$f(x)=(x+1)^2(x-2).$$
