# How do you find the product of (3a-4)(3a+4)?

The result is $9 {a}^{2} - 16$
The problem is an example of a notable product: "the sum multiplied by the diference is equal to the difference of squares", that is to say: (a+b)·(a-b) = a^2-b^2.
(3a-4)·(3a+4) = (3a)^2-(4)^2 = 9a^2 - 16.