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?
 A: 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:
$$x^3-6x+4=(x-2)(x^2+2x-2)$$
A: 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. 
A: 
Note: I understand that there is already an accepted answer for this question, so this answer may be useless, but regardless, I'm still posting this to spread knowledge!

A simple way to factorize depressed cubic polynomials of the form$$x^3+Ax+B=0\tag1$$
Is to first move all the constants to the RHS, so $(1)$ becomes$$x^3+Ax=-B\tag2$$
Now, find two factors of $B$ such that one fact minus the square of the other factor is $A$. We'll call them $a,b$ so$$\begin{align*} & a-b^2=A\tag3\\ & ab=-B\tag4\end{align*}$$
Multiply $(2)$ by $x$, add $b^2x^2$ to both sides and complete the square. Solving should give you a value of $x$ and allow you to factor $(1)$ by Synthetic Division.


Examples:



  
*
  
*Solve $x^3-6x+4=0$ (your question)
  

Moving $4$ to the RHS and observing its factors, we have $-2,2$ as $a,b$ since$$-2-2^2=A\\-2\cdot2=-4$$Therefore, we have the following:$$x^4-6x^2=-2\cdot2x$$$$x^4-6x^2+4x^2=4x^2-4x$$$$x^4-2x^2=4x^2-4x$$$$x^4-2x^2+1=4x^2-4x+1\implies(x^2-1)^2=(2x-1)^2$$$$x^2=2x\implies x=2$$
Note that we do have to consider the negative case when square rooting, but they lead to the same pair of answers. So it's pointless.


  
*Solving $x^3+16x=455$
  

A factor of $455$ works, namely when $a=65,b=7$.$$65-7^2=16$$$$65\cdot7=455$$
Therefore,$$x^4+16x^2=65\cdot7x$$$$x^4+65x^2=49x^2+455x$$$$\left(x^2+\dfrac {65}{2}\right)^2=\left(7x+\dfrac {65}{2}\right)^2$$$$x=7$$
A: The rational root theorem gives a list of all possible rational roots of a polynomial with integer coefficients that have a given leading coefficient and a given constant coefficient. In this case, the leading coefficient is $1$ and the constant coefficient is $4.$ The theorem tells us that all rational roots are in the set $\left\{ \pm\dfrac 1 1, \pm\dfrac 2 1, \pm \dfrac 4 1 \right\},$ the numerator being in this the only divisor of the leading coefficient $1$ and the denominators being the divisors of the constant coefficient $4$. That doesn't mean there are rational roots; it only means there are not any that don't belong to this set. There are only six members of this set, so it's easy to plug in all of them and see if you get $0$. When you plug in $2$, you get $0$, so there's your factorization.
A: If $P$ is a polynomial with real coefficients and if $a\in\mathbb{R}$ is a root, which means that $P(a)=0$, then there exists a real polynomial $Q$ such that $\forall x\in\mathbb{R},\quad P(x)=(x-a)\,Q(x)$.
On this case, you can see by inspection, that $P(2)=0$.
It remains to find real constants $A,B,C$ such that :
$$\forall x\in\mathbb{R},\quad x^3-6x+4=(x-2)(Ax^2+Bx+C)$$
Identification of coefficients leads to $A=1$, $-2C=4$ and, for example, $A-2B=0$ (equating the coeffts of $x^2$ in both sides).
