How do you factor completely $49{{x}^{2}}-9$?

Hint: In this question we have been given a polynomial equation of degree $2$. The given expression is not in the form of a quadratic equation hence there is no direct way to factorize it. Therefore, to factorize the expression we will convert the expression into the format of ${{a}^{2}}-{{b}^{2}}$ such that it can be expanded using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to get the required solution.

Complete step by step solution:
We have the given expression as:
Now we can see in the expression that the terms $49{{x}^{2}}$ and $9$ are present, which have square roots therefore, we can utilize their square roots to simplify the expression.
We know that $49{{x}^{2}}={{\left( 7x \right)}^{2}}$ and $9={{3}^{2}}$ therefore, on substituting it in the expression, we get:
$\Rightarrow {{\left( 7x \right)}^{2}}-{{\left( 3 \right)}^{2}}$
Now the above expression is in the form of ${{a}^{2}}-{{b}^{2}}$ therefore, we can expand it using the expansion formula.
On using the formula, we get:
\[\Rightarrow \left( 7x-3 \right)\left( 7x+3 \right)\], which is the required solution.

Note:
To check whether the solution is correct, we will multiply the factors, if it gives the same term, then the solution is correct:
On multiplying, we get:
\[\Rightarrow 7x\times 7x+7x\times 3-3\times 7x-3\times 3\]
On multiplying the terms, we get:
\[\Rightarrow 49{{x}^{2}}+21x-21x-9\]
We know that similar terms with opposite signs get cancelled therefore, on cancelling, we get:
\[\Rightarrow 49{{x}^{2}}-9\], which is the original expression therefore, the solution is correct.
In the above question we have a polynomial equation of degree $2$. Even though the polynomial equation has a degree $2$, it is not a complete quadratic equation because it has the value of coefficient of $x$ as $0$.
It is to be remembered that factors are the digits which make up a number, a factor should be indivisible by any number except $1$ and it should not be in the form of a number which is raised to a power.