How do you factor : $3{{x}^{3}}-9{{x}^{2}}$ ?

Hint: We can see that the equation given in the question is a cubic equation where the coefficient of x is 0 and the constant term is 0. So we can take ${{x}^{2}}$ common to factorize the equation.

Complete step by step answer:
 The given equation which we have to simplify is $3{{x}^{3}}-9{{x}^{2}}$
We can see that in the above cubic equation the coefficient of 0 and the constant term is 0, so we can take ${{x}^{2}}$ common form the equation
So by taking ${{x}^{2}}$ common we get
$\Rightarrow 3{{x}^{3}}-9{{x}^{2}}={{x}^{2}}\left( 3x-9 \right)$ ..eq1
Now we can see that in the term $3x-9$ , we can take 3 common
So we can write $\Rightarrow 3x-9=3\left( x-3 \right)$
So by replacing 3x – 9 with $3\left( x-3 \right)$ in eq1 we get
$\Rightarrow 3{{x}^{3}}-9{{x}^{2}}=3{{x}^{2}}\left( x-3 \right)$
$3{{x}^{2}}\left( x-3 \right)$ is the factored form of the equation $3{{x}^{3}}-9{{x}^{2}}$

Note:
If we are given a polynomial equation of degree n then the factored form of the polynomial equation is $a\left( x-{{\alpha }_{1}} \right)\left( x-{{\alpha }_{2}} \right)........\left( x-{{\alpha }_{n}} \right)$ where a is the coefficient of ${{x}^{n}}$ and ${{\alpha }_{1}}$ , ${{\alpha }_{2}}$ , ….., ${{\alpha }_{n}}$ are roots of the equation. Sometimes the roots of the polynomial equation can be repetitive. Let’s find the roots of the equation given in the question
The degree of the equation $3{{x}^{3}}-9{{x}^{2}}$ is equal to 3. So all the roots of equation $3{{x}^{3}}-9{{x}^{2}}$ are 0, 0, and 3 . We can see the root 0 is repetitive here. The coefficient of ${{x}^{3}}$ in the equation $3{{x}^{3}}-9{{x}^{2}}$ is 3. So the factored form of the equation $3{{x}^{3}}-9{{x}^{2}}$ is $3\left( x-0 \right)\left( x-0 \right)\left( x-3 \right)$ which is $3{{x}^{2}}\left( x-3 \right)$