How do you multiply $(2r^2+7s^2)(2r^2-7s^2)$ ?

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Answer 1 · 2016-08-14T20:53:14

$4r^4 -49s^4$

We could use the usual method of binomial times binomial, but we should notice that this is the answer to the difference of squares.

$(x+y)(x-y) = x^2 - y^2$

Therefore in this case:

$(2r^2+7s^2)(2r^2-7s^2) = 4r^4 -49s^4$

How do you multiply (2r^2+7s^2)(2r^2-7s^2)(2r^2+7s^2)(2r^2-7s^2)?