How do you factor ${{x}^{3}}-13{{x}^{2}}-30x$ completely?

Hint: In this problem we need to find the factor of ${{x}^{3}}-13{{x}^{2}}-30x$. We can observe that the given equation is a cubic equation. Now we will equate the given equation to zero and take the common term which is $x$ as common from the equation. Now we will equate each term in the obtained equation to zero. Here we will get one factor and a quadratic equation. For quadratic equation we will split the middle term or the coefficient of $x$ such that $ax=bx+cx$ where the product of $b$, $c$ should be equal to the product of ${{x}^{2}}$ coefficient and constant in the quadratic equation. After getting the factors of the quadratic equation we will write the factors of the given equation.

Complete step-by-step answer:
Given that, ${{x}^{3}}-13{{x}^{2}}-30x$.
Equating the above equation to zero, then we will get
${{x}^{3}}-13{{x}^{2}}-30x=0$
Taking $x$ as common in the above equation, then we will have
$x\left( {{x}^{2}}-13x-30 \right)=0$
Equating each term individually to zero, then we will get $x=0$, ${{x}^{2}}-13x-30=0$. Here $x=0$ represents that $x$ is a factor of the given equation. For the remaining factors we need to factorize the equation ${{x}^{2}}-13x-30=0$.
In the above quadratic equation
The first term is, ${{x}^{2}}$ its coefficient is 1.
The middle term is, $13{{x}^{2}}$its coefficient is -13
The last term is, constant is $-30$
Now multiply the coefficient of the first term by constant
$1\left( -30 \right)=-30$
Finding the two factors of $-30$ whose sum equal the coefficient of the middle term which is $-13$.
We have factors for the $30$ are $1,2,15,30$. So
$\begin{align}
  & -15+2=-13 \\
 & -15\times 2=-30 \\
\end{align}$
From the above values we can write the quadratic equation as
${{x}^{2}}-15x+2x-30=0$
Taking $x$ as common from ${{x}^{2}}-15x$ and $2$ as common from $2x-30$, then we will get
$x\left( x-15 \right)+2\left( x-15 \right)=0$
Again, taking $x-15$ as common from the above equation, then we will get
$\left( x+2 \right)\left( x-15 \right)=0$
Hence the factors of the given equation are $x$, $x+2$, $x-15$.

Note: In the above solution we have splitted the middle term into two terms. We can also find the roots of the roots of the quadratic equation from the well know formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.After getting the roots of the equation as $x=\alpha $, $x=\beta $ we can write the factors of the quadratic equation as $x-\alpha $, $x-\beta $.