How do you factor \[16{x^2} - 25\] ?

Hint: To find the factor of \[16{x^2} - 25\], make the equation of the form of whole square and then use the identity, ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ , which will give us the required factorization for the given equation.

Complete Step by Step by Solution:
We have been given the equation, \[16{x^2} - 25\]. So, first, we have to simplify the given equation so, that we can use the suitable identity to solve the given equation.
We have given with the $16{x^2}$, $x$ already has the power of 2. When we do the square root of 16, we get the answer as 4. So, we can rewrite $16{x^2}$ as ${\left( 4 \right)^2}{\left( x \right)^2}$ . Then, by putting the brackets in both $4$ and $x$, we can write as, ${\left( {4x} \right)^2}$.
Another number in the equation is 25. If we do the square root of 25, then we’ll get the square root of 25 is 5. Then, by putting the brackets around 5, we can write it as, ${\left( 5 \right)^2}$.
Now, the equation given in the question can be rewritten as –
$ \Rightarrow {\left( {4x} \right)^2} - {\left( 5 \right)^2} \cdots \left\{ 1 \right\}$
From the above, we can conclude that the above polynomial is the difference of squares and there is the formula to factor these type of polynomials –
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
By using the above identity in the equation {1}, we get –
$ \Rightarrow {\left( {4x} \right)^2} - {\left( 5 \right)^2} = \left( {4x - 5} \right)\left( {4x + 5} \right)$

Hence, the factorization of the equation \[16{x^2} - 25\] is $\left( {4x - 5} \right)\left( {4x + 5} \right)$.

Note:
Using the factors of \[16{x^2} - 25\], we can also find the values of $x$.
We have to put each factor equal to 0 and then, by using the transposition method we can shift the term to another side and we will get the values of $x$ -
$ \Rightarrow 4x - 5 = 0,4x + 5 = 0$
Shifting the terms to another side, we get –
$ \Rightarrow x = \dfrac{5}{4},x = \dfrac{{ - 5}}{4}$
Hence, these are the required values of $x$.