Factorise the expression $4{{p}^{2}}-9{{q}^{2}}$

Hint: We first try to explain the concept of factorisation and the ways a factorisation of a polynomial can be done. We use the identity theorem of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factor the given polynomial $4{{p}^{2}}-9{{q}^{2}}$. We assume the values of $a=2p;b=3q$. The final multiplied linear polynomials are the solution of the problem.

Complete step-by-step solution:
The main condition of factorisation is to break the given number or function or polynomial into multiple of basic primary numbers or polynomials.
For the process of factorisation, we use the concept of common elements or identities to convert into multiplication form.
For the factorisation of the given quadratic polynomial $4{{p}^{2}}-9{{q}^{2}}$, we apply the factorisation identity of difference of two squares as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We get $4{{p}^{2}}-9{{q}^{2}}={{\left( 2p \right)}^{2}}-{{\left( 3q \right)}^{2}}$. We put the value of $a=2p;b=3q$.
Factorisation of the polynomial gives us
$4{{p}^{2}}-9{{q}^{2}}={{\left( 2p \right)}^{2}}-{{\left( 3q \right)}^{2}}=\left( 2p+3q \right)\left( 2p-3q \right)$.
These two multiplied linear polynomials can’t be broken any more.
Therefore, the final factorisation of $4{{p}^{2}}-9{{q}^{2}}$ is $\left( 2p+3q \right)\left( 2p-3q \right)$.

Note: we can verify the factorisation of the equation assuming arbitrary values for $p$ and$q$ and putting them on $4{{p}^{2}}-9{{q}^{2}}=\left( 2p+3q \right)\left( 2p-3q \right)$.
We assume $p=4$ and $q=1$. We put the values on the left side of the equation and get
$4{{p}^{2}}-9{{q}^{2}}=4\times {{4}^{2}}-9\times {{1}^{2}}=64-9=55$.
We now put the values on the right side of the equation and get
$\left( 2p+3q \right)\left( 2p-3q \right)=\left( 2\times 4+3\times 1 \right)\left( 2\times 4-3\times 1 \right)=11\times 5=55$.
Thus verified $4{{p}^{2}}-9{{q}^{2}}=\left( 2p+3q \right)\left( 2p-3q \right)$.