How do you factor completely $12{{x}^{2}}-27$?

Hint: We know 3 is the common factor of 12 and 27, so we can take 3 common from the equation. We know the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can apply this formula to solve the problem very easily.

Complete step by step answer:
The given equation which we have to factor is $12{{x}^{2}}-27$
We will take 3 common from the equation because we know 3 is the common factor of 12 and 27
$\Rightarrow 12{{x}^{2}}-27=3\left( 4{{x}^{2}}-9 \right)$
$4{{x}^{2}}$ is the square of 2x and 9 is square of 3
So we can write $3\left( 4{{x}^{2}}-9 \right)$ as $3\left( {{\left( 2x \right)}^{2}}-{{3}^{2}} \right)$
We can apply the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above equation if we assume a is equal to 2x and b is equal to 3
Then $3\left( 4{{x}^{2}}-9 \right)$ is equal to $3\left( 2x+3 \right)\left( 2x-3 \right)$
Now we can say $12{{x}^{2}}-27$ is equal to $3\left( 2x+3 \right)\left( 2x-3 \right)$

Note:
We can solve the problem by the method we use to factorize quadratic equations.
 if we compare the given equation in the problem statement after taking 3 common to quadratic equation $a{{x}^{2}}+bx+c$ we get a is equal to 4 , b is equal to 0 and c is equal to -9
So we will find 2 number such that their product is equal to product of a and c which is equal to -36, and their sum is equal to 0
Let’s make a list of integer pair with product equals to -36 and we will take out the one with sum 0
(1,-36), (2, -18), (3,-12), (4,-9), (6,-6), (9,-4), (-3, 12), (-2, 18), (-1,36)
The pair (6,-6) has sum equals 0
$\Rightarrow 3\left( 4{{x}^{2}}-9 \right)=3\left( 4{{x}^{2}}+6x-6x-9 \right)$
Taking 2x common in first half and -3 common in second half of the equation we get
$\Rightarrow 3\left( 4{{x}^{2}}-9 \right)=3\left( 2x\left( 2x+3 \right)-3\left( 2x+3 \right) \right)$
Taking $2x+3$ common
$\Rightarrow 3\left( 4{{x}^{2}}-9 \right)=3\left( 2x+3 \right)\left( 2x-3 \right)$