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

1 Answer
Jose Luis F.
Apr 10, 2015

The result is 9a^2 - 169a2−16

The reason is the following:

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(a+b)⋅(a−b)=a2−b2.
By applying this to our question, we obtain that:
(3a-4)·(3a+4) = (3a)^2-(4)^2 = 9a^2 - 16(3a−4)⋅(3a+4)=(3a)2−(4)2=9a2−16.

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