How do you factor $5{{x}^{3}}-20{{x}^{2}}$ ? \[\]

Hint: We recall the Euclidean division of polynomials where we say that the divisor polynomial is a factor polynomial when the remainder polynomial is zero. We take 5 common from both the terms of given polynomial $5{{x}^{3}}-20x$ and to get $5x\left( {{x}^{2}}-4 \right)$. We then use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=x,b=2$ to factorize completely .

Complete step by step answer:
We know that when we divide a divided polynomial $p\left( x \right)$ with degree $n$ by some divisor polynomial $d\left( x \right)$ with degree $m\le n$ then we get the quotient polynomial $q\left( x \right)$ of degree $n-m$ and the remainder polynomial as $r\left( x \right)$ .We use Euclidean division formula and can write as
\[ p\left( x \right)=d\left( x \right)q\left( x \right)+r\left( x \right)\]
We also know that if the remainder polynomial is zero $\left( r\left( x \right)=0 \right)$ then we call $d\left( x \right),q\left( x \right)$ factor polynomials of $p\left( x \right)$ or simply factors of $p\left( x \right)$. If ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{k}}\left( x \right)$ are $k$ factors of $p\left( x \right)$ then we say $p\left( x \right)={{p}_{1}}\left( x \right){{p}_{2}}\left( x \right)...{{p}_{k}}\left( x \right)$ is factored completely if none of the factors ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{k}}\left( x \right)$ can be factored further. We are given the following polynomial in the question
\[ 5{{x}^{3}}-20x\]
We can write the above polynomial as
\[\begin{align}
  & \Rightarrow 5{{x}^{3}}-5\times 4\times x \\
 & \Rightarrow 5x\times {{x}^{2}}-5\times 4x \\
\end{align}\]
We take $5x$ common from both the terms in the above step to have;
\[\Rightarrow 5x\left( {{x}^{2}}-4 \right)\]
We see that expression in the bracket ${{x}^{2}}-4$ is in the form of ${{a}^{2}}-{{b}^{2}}$ because ${{x}^{2}}-4={{\left( x \right)}^{2}}-{{\left( 2 \right)}^{2}}$. We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=x,b=2$ in the above step to have-
\[\begin{align}
  & \Rightarrow 5x\left( x+2 \right)\left( x-2 \right) \\
 & \Rightarrow 5{{x}^{3}}-20{{x}^{2}}=5x\left( x+2 \right)\left( x-2 \right) \\
\end{align}\]
The above factorization is the complete factorization since $5,x,x+2,x-2$ cannot be factored further.

Note:
We note that the highest power on the variable is called degree of the polynomial. If degree is 1 we call the polynomial a linear polynomial. Here in the factorization of $5{{x}^{3}}-20x$ the obtained factors $x,x+2,x-2$ are linear factors. We should remember linear factors cannot be factored further. We can alternatively use factor theorem to factorize which states that $x-a$ is a factor of $p\left( x \right)$ if $p\left( a \right)=0$. We guess a zero $x=a$ of $p\left( x \right)$ and then divide $p\left( x \right)$ by $x-a$ to get the other factor.