How do you factor $ 4{x^2} - 9 $ ?

Hint: In this question we need to find the factor for $ 4{x^2} - 9 $ . Here, we will rewrite the given equation, as the numbers are perfect squares. Then, it will be converted into the form of difference of squares. Here, we will apply the formula of difference of squares. And, substitute the value of $ a $ and $ b $ , by which we will get the required factor of the given.

Complete step-by-step answer:
Now, we need to find the factor for $ 4{x^2} - 9 $ .
We can see here that both $ 4 $ and $ 9 $ are perfect squares.
Therefore, we can rewrite $ 4{x^2} - 9 $ as,
 $ {\left( {2x} \right)^2} - {\left( 3 \right)^2} $
Now, let us factor using the difference of squares formula,
 $ {a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) $
Here, $ a = 2x $ and $ b = 3 $
Now, by substituting the values, we have,
 $ {\left( {2x} \right)^2} - {\left( 3 \right)^2} = \left( {2x - 3} \right)\left( {2x + 3} \right) $
Hence, the factors of $ 4{x^2} - 9 $ are $ \left( {2x - 3} \right) $ and $ \left( {2x + 3} \right) $ .
So, the correct answer is “ $ \left( {2x - 3} \right) $ and $ \left( {2x + 3} \right) $ ”.

Note: It is important to note here that when we are facing a quadratic equation, first we will group the polynomial into two sections. Find what is common in both sections. Then, factor the commonalities out of the two terms, if each of the two terms contains the same factor, which is the required solution. However here we have used the difference of squares formula as the numbers are in perfect square, we can’t use this method for all quadratic equations. A quadratic polynomial is a polynomial of the form, $ a{x^2} + bx + c $ where $ a $ is non-zero. Normally, the quadratic equation can be solved in other ways too, which depends on the polynomial given. However, with a little practice solving the quadratic equation becomes a piece of cake. Also, being aware of the formulas helps us solve these kinds of problems.