How do you find the product of $(3+7v^2)(3-7v^2)$ ?

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Answer 1 · 2016-07-22T15:53:53

$9-49v^4$

This is a binomial $xx$ a binomial.

Each term in the first bracket must be multiplied by each term in the second bracket. This will give $2xx2 = 4$ terms.

$(3+7v^2)(3-7v^2)$

= $9 -21v^2 +21v^2 -49v^4$

The middle terms are additive inverses, leaving us with..

$9-49v^4$

We could also have recognised this as the factors of the Difference of two squares (the brackets are the same apart from the signs),

and written down the two squares immediately.

How do you find the product of (3+7v^2)(3-7v^2)(3+7v^2)(3-7v^2)?