Solve: \[4{{a}^{2}}-9{{b}^{2}}-2a-\text{ }3b\]

Hint: We know that the number 4 is the square of 2 and the number 9 is the square of 3. So, what we will do is first we will write $4{{a}^{2}}={{(2a)}^{2}}$ and $9{{b}^{2}}={{(3b)}^{2}}$. After that we will use the identity \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] in \[4{{a}^{2}}-9{{b}^{2}}\], and then will simplify the expression to obtain the solution.

Complete step-by-step answer:
Before solving this question, we must know about square and square roots of numbers.
A square number is the result when a number is multiplied by itself. For example, 25 is a square number because \[5\times 5\] is 25. 100 is also a square number because it's\[{{\left( 10 \right)}^{2}}\].
The square root of a number is a special value that, when used in a multiplication two times, gives that number.
Example: \[2\times 2=4\] , so the square root of 4 is 2, \[3\times 3=9\] , so the cube root of 9 is 3.

Let us now solve this question.
As mentioned in the hint provided above, the number 4 is the square of 2 and the number 9 is the square of 3.
Therefore, \[4{{a}^{2}}-9{{b}^{2}}-2a-\text{ }3b\] can also be written as follows:-
\[\left[ \left( 2a{{)}^{2}}-{{\left( 3b \right)}^{2}}\left] \left( 2a+3b \right)= \right[\left( 2a+3b \right)\left( 2a-3b \right) \right]\left( 2a+3b \right) \right)\]
\[=\left( 2a+3b \right)\left( 2a-3b-1 \right)\]
So, as explained above, \[4{{a}^{2}}-9{{b}^{2}}-2a-3b\] can also be written as (2a + 3b) (2a – 3b – 1).
Therefore, the answer of the question is (2a + 3b) (2a – 3b – 1).

Note: The identity used in the solution of this question is \[{{a}^{2}}{{b}^{2}}=\left( a+b \right)\left( ab \right)\] where, a = 2a and b = 3b. There are some other identities also which are very useful in such questions such as
a 2 + 2ab + b 2 = (a + b) 2 , a 2 – 2ab + b 2 = (a – b) 2, x 2 + (a + b) x + ab = (x + a) (x + b). Try not to make any calculation errors while solving the question.