How do you factor completely \[{{x}^{3}}-4{{x}^{2}}-2x+8\]?

Hint: In order to solve this question, firstly find a root \[x=a\] which fits the equation \[{{x}^{3}}-4{{x}^{2}}-2x+8=0\] perfectly. Then, just divide the factor \[\left( x-a \right)\] with the equation that we need to factorize and we get a quadratic equation as the remainder. The quadratic equation is factored then.

Complete step-by-step solution:
 Let us start solving the question by taking the equation,
\[{{x}^{3}}-4{{x}^{2}}-2x+8=0\]
Now, let us think of a root that completely satisfies this equation,
If
 \[\begin{align}
  & x=4 \\
 & \Rightarrow {{4}^{3}}-4\centerdot {{4}^{2}}-2\left( 4 \right)+8=64-64-8+8=0 \\
\end{align}\]
So, we can say that the factor \[\left( x-4 \right)\] satisfies the equation, now dividing the equation\[{{x}^{3}}-4{{x}^{2}}-2x+8\] with \[\left( x-4 \right)\], we get
\[\left( x-4 \right){{\left| \!{\overline {\,
 \begin{align}
  & {{x}^{3}}-4{{x}^{2}}-2x+8 \\
 & _{-}{{x}^{3}}{{-}_{+}}4{{x}^{2}}\downarrow +\downarrow \\
 & 0+0 \\
 & -2x+8 \\
 & _{+}-2x{{+}_{-}}8 \\
 & =0 \\
\end{align} \,}} \right. }^{{{x}^{2}}-2}}\]
Therefore, we can say that
\[\left( x-4 \right)\left( {{x}^{2}}-2 \right)={{x}^{3}}-4{{x}^{2}}-2x+8\]
So, now, we just need to make the further factors for the factor \[\left( {{x}^{2}}-2 \right)\]
\[\begin{align}
  & \left( {{x}^{2}}-2 \right) \\
 & \Rightarrow \left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right) \\
\end{align}\]
Thus, we can write that the factors for the equation\[{{x}^{3}}-4{{x}^{2}}-2x+8\] are
\[\left( x-4 \right)\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)\]
Hence the equation has been factored.

Note: The given equation is a cubic equation, that is why we get three factors. However, as the factor \[\left( x-4 \right)\] is repeated twice, its square has been taken but the degree of the equation still remains the same i.e. \[3\]. The factors can be verified by further multiplying them to get the same equation as before.