Factorize $x^3-3x+2$ How can I factorize $x^3-3x+2$ ?
The answer that I got on the internet is $x^3-2x^2+x+2x^2-4x+2$=$(x-1)^2(x+2)$ 
It would be nice if anyone could also tell what these type of equations are called and where can I learn more?
 A: They are called cubic functions / cubic equations. A closed formula for the solutions exists but it is quite ugly so the common method to factorize the term is to guess one root $x_0$ and then do long division by $(x-x_0)$.
So the method one would be to use the formula on
$x^3-3x+2=0$ and find that the roots are $1,1,-2$ and you are done. The trivial method is to guess $x_0=1$ and use the long division
A: Just in case you're wondering what is meant by a rational "root": a root for a given polynomial can be thought of as a value $r$ such that when you set $x = r$, the value of the polynomial evaluates to zero.  Some people refer to the roots as the "zeros" of a polynomial.
So, I'll use $P(x)$ to denote the equation $x^3-3x+2$; that is, we have
$$P(x) = x^3 - 3x + 2 = (x-1)^2(x+2)$$
Now, when is $P(x) = 0$?
When $(x-1)=0$ and/or when $(x+2)=0$, because when that happens, then $$P(x) = (0)(x+1) =0\ \text{or}\ P(x) =(x-1)^2(0) = 0$$
Now we simply solve for $x$ to find the roots:
$(x-1) = 0\implies x=1$, so $1$ is a root of $P(x)$, and
$(x+2) = 0 \implies x = -2$, so $-2$ is a root of $P(x)$,
and, voila, you have the roots, as given in the first answer.
In terms of learning how to factor a polynomial, the best advice I have to offer is practice!
A: If the roots are rational you can get them by finding all positive and negative factors of the last digit and divide them by the same of the first digit. Then test them $\pm2$ and $\pm1$. We get 1 and $-2=x$. Then divide by their respective polynomials to see which one has order 2. Solved.
