What is the step by step solution to factor $x^3 -3x -2$ My younger brother has this problem in school, and I don't know how to explain to him the solution other that to use intuition. Is there algorithmic step by step solution to solve these kinds of problems?
the result is $(x+1)^2 (x-2)$ 
 A: This is a depressed cubic i.e. something of the form $\color{red}{x^3 + \beta_1 x + \beta_0 = 0}$. I will here show the general algorithmic approach to solve it. It usually involves two substitutions:


*

*First substitute $\color{red}{x = y - \frac{\beta_1}{3y}}$. This results in
$$ 0 = y^3 + \beta_0 - \frac{\beta_1^3}{27y^3}=y^3 -2 + \frac{27}{27y^3}$$

*Then you multiply by $y^3$ and subsitute $\color{red}{z = y^3}$ so you get
$$0 = z^2 -2z+1 = (z-1)^2$$
So $z=1$. 

*Now we need to invert the substitutions:
Because $z =y^3$ we get $y=1$ (or $e^{2\pi k/3}$ for $k=0,1,2$ if you want to include complex solutions, but one solution is sufficient to reduce the degree). 
And because $x = y - \frac{-3}{3y}$ we get $x=2$ as a possible solution. If we plug this into our depressed cubic, we see that $x=2$ does indeed satisfy the equation. This means we can factor out $(x-2)$ via polynomial long division and then factor the remaining quadratic term as usual.
A: $$x^3-3x-2=x^3-2x^2+2x^2-4x+x-2=(x-2)(x^2+2x+1)=(x-2)(x+1)^2$$
A: The standard way is to check the divisors of $-2$ for a soluton of $x^3-3x-2=0$.
These are $\pm 2$ and $\pm 1$.
You see, that for $x=2$ you get $8-6-2=0$ and then calculate $(x^3-3x-2)\div (x-2)=x^2+2x+1=(x+1)^2$ which gives the factorisation $x^3-3x-2=(x+1)^2(x-2)$
