How do you factor ${{x}^{3}}-6{{x}^{2}}+5x$ ?

Hint: To factor the given expression ${{x}^{3}}-6{{x}^{2}}+5x$, we will first of all take $x$ as common from this expression. After taking $x$ as common, we will get the quadratic expression inside the bracket. And further we can factorize quadratic expression by factorization method. Let us suppose that we have a quadratic expression say ${{x}^{2}}+bx+c$ then to factorize it we will multiply the coefficient of ${{x}^{2}}$ and the constant and then factorize this multiplication in such a way that addition or subtraction of this multiplication is equal to b. Then it will be easier for you to further factorize the quadratic expression.

Complete step by step answer:
The expression that we have to factorize is as follows:
${{x}^{3}}-6{{x}^{2}}+5x$
Taking $x$ as common from the above expression we get,
$x\left( {{x}^{2}}-6x+5 \right)$
Now, as you can see that the expression written inside is the quadratic expression $\left( {{x}^{2}}-6x+5 \right)$ and we are going to factorize this quadratic expression in the following way:
We are going to write the factors for 5 which is equal to:
$5=5\times 1$
Now, adding the two factors 5 and 1 will give us 6 so we can write $\left( 5+1 \right)$ in place of 6 in the quadratic equation and we get,
${{x}^{2}}-\left( 5+1 \right)x+5$
Multiplying one by one the elements written inside the bracket with $x$ we get,
${{x}^{2}}-5x-x+5$
 Taking $x$ as common from ${{x}^{2}}-5x$ and -1 from $-x+5$ we get,
$x\left( x-5 \right)-1\left( x-5 \right)$
Now, in the above, as you can see that we can take $\left( x-5 \right)$ as common then we get,
$\left( x-5 \right)\left( x-1 \right)$
Hence, we have factorize the quadratic expression $\left( {{x}^{2}}-6x+5 \right)$ to $\left( x-5 \right)\left( x-1 \right)$.
Hence, we have factorized the given expression as $x\left( x-5 \right)\left( x-1 \right)$.

Note:
Instead of finding the factors of the quadratic expression in the way we have done above, we can factorize the quadratic equation $\left( {{x}^{2}}-6x+5 \right)$ by finding the roots of this quadratic expression using Shree Dharacharya rule.
The roots of the quadratic equation $\left( a{{x}^{2}}+bx+c=0 \right)$ using Shree Dharacharya rule is as follows:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Using the above Shree Dharacharya rule in finding the roots of the quadratic equation $\left( {{x}^{2}}-6x+5=0 \right)$ we get,
$\begin{align}
  & x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 1 \right)\left( 5 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{6\pm \sqrt{36-20}}{2} \\
 & \Rightarrow x=\dfrac{6\pm \sqrt{16}}{2} \\
 & \Rightarrow x=\dfrac{6\pm 4}{2} \\
\end{align}$
Taking the root with positive sign we get,
$\begin{align}
  & x=\dfrac{6+4}{2} \\
 & \Rightarrow x=\dfrac{10}{2}=5 \\
\end{align}$
Taking the root with negative sign we get,
$\begin{align}
  & x=\dfrac{6-4}{2} \\
 & \Rightarrow x=\dfrac{2}{2}=1 \\
\end{align}$
Hence, we have found the roots of this quadratic equation as $\left( {{x}^{2}}-6x+5=0 \right)$ as $\left( x=5,x=1 \right)$.
Now, rearranging the roots of the quadratic equation we get,
$\begin{align}
  & x-5=0; \\
 & x-1=0 \\
\end{align}$
Now, multiplying the above two equations we will get the above quadratic equation $\left( {{x}^{2}}-6x+5=0 \right)$.
$\left( x-5 \right)\left( x-1 \right)=0$
Or, we can write the factorization of quadratic expression $\left( {{x}^{2}}-6x+5 \right)$ as $\left( x-5 \right)\left( x-1 \right)$.