How to factorize this cubic equation? In one of the mathematics book, the author factorized following term
$$x^3 - 6x + 4 = 0$$
to
$$( x - 2) ( x^2 + 2x -2 ) = 0.$$
How did he do it?

There is a neat trick called the rational roots theorem.  All we have to do is factor the first and last numbers, put them over a fraction, and take $\pm$.  This gives us the following possible rational roots:
$$x\stackrel?=\pm1,\pm2,\pm4$$
due to the factorization of $4$.  Checking these, it is clear $x=2$ is the only rational root, since
$$\begin{align}0&\ne(+1)^3-6(+1)+4\\0&\ne(-1)^3-6(-1)+4\\\color{#4488dd}0&=\color{#4488dd}{(+2)^3-6(+2)+4}\\0&\ne(-2)^3-6(-2)+4\\0&\ne(+4)^3-6(+4)+4\\0&\ne(-4)^3-6(-4)+4\end{align}$$
leaving us with
$$x^3-6x+4=(x-2)(\dots)$$
We can find the remainder through synthetic division:
$$\begin{array}{c|c c}2&1&0&-6&4\\&\downarrow&2&4&-4\\&\hline1&2&-2&0\end{array}$$
which gives us our factorization:
$$x^3-6x+4=(x-2)(x^2+2x-2)$$
