How to factorize this cubic equation?

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:

$$x^3-6x+4=(x-2)(x^2+2x-2)$$

Since you do not know the Rational Root Test, let's consider a simpler case: the Integer Root Test.

If $\,f(x)= x^3+6x+4\,$ has an integer root $\,x=n\,$ then $\,n^3+6n+4 = 0\,$ so $\,(n^2+6)\,\color{#c00}{n = -4},\,$ hence $\,\color{#c00}{n\ \ {\rm divides}\ \ 4}.\,$ Testing all the divisors of $4$ shows that $2$ is root, $ $ hence $\,x-2\,$ is a factor of $f$ by the Factor Theorem. The cofactor $\,f/(x-2)\,$ is computable by the Polynomial (long) Division algorithm (or even by undetermined coefficients).

Remark $\ $ This is a very special case of general relations between the factorization of polynomials and the factorizations of their values. For example, one can derive relations between primality and compositeness of polynomials based on the same properties of their values. For example, since $\ 9^4\!+8\ $ is prime so too is $\, x^4+8\,$ by Cohn's irreducibility test. See this answer and its links for some of these beautiful ideas of Bernoulli, Kronecker, and Schubert.