How do you factor $(25{x^2} - 9)$?

Hint: First write $25\;$in terms of exponential form and then group it with the $x$ variable. Now use the ${a^2} - {b^2}$ formula to split the expression into the sum of products form which is also known as the factors for the given expression.

Formula used:
$\Rightarrow$${a^2} - {b^2} = (a + b)(a - b)$

Complete step-by-step answer:
First, write $25\;$in exponential form, in terms of squares so that it can easily be grouped with the variable and can be collectively written together in square form.
$ \Rightarrow {5^2}{x^2} - 9$
Now, write $9$ also in the square form so that it is later useful for us to use the distributive law.
$ \Rightarrow {5^2}{x^2} - {3^2}$
Now, group $5$ and the variable $x$ to get the form of the formula of the distributive law.
$ \Rightarrow {(5x)^2} - {(3)^2}$
This is now in the form of ${a^2} - {b^2}$ which is equal to $(a + b)(a - b)$
To prove this formula, just multiply the linear line equations and checkout.
Here,$a = 5x;b = 3$
On substitution we finally get,
$ \Rightarrow (5x + 3)(5x - 3)$

$\therefore $The given expression$(25{x^2} - 9)$ factories into $(5x + 3)(5x - 3)$

Additional information: The process of factorization is reverse multiplication. In the above question, we have multiplied two linear line equations to get a quadratic equation (a polynomial of degree $2$) expression using the distributive law. The name quadratic comes from “quad” meaning square because the variable gets squared. Also known as a polynomial of degree $2$.

Note:
This can also be solved by another method, which is the sum-product pattern, where the sum of roots is the value of the degree $1$ term of $x$ and the product of roots is the value of degree $2$term and degree $0$ term multiplied together.
Product of the roots must be $25 \times 9 = 225$
And the sum of the roots must be $0$
We can write $25{x^2} - 9 = 25{x^2} + 15x - 15x - 9$
On solving it further,
$ \Rightarrow 5x(5x + 3) - 3(5x + 3)$
On factoring,
$ \Rightarrow (5x + 3)(5x - 3)$