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?


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,$$x^4+16x^2=65\cdot7x$$$$x^4+65x^2=49x^2+455x$$$$\left(x^2+\dfrac {65}{2}\right)^2=\left(7x+\dfrac {65}{2}\right)^2$$$$x=7$$
