Question

Determine whether 81 − 49n4 is a difference of two squares. If so, factor it. If not, explain why.
A (9 − 7n^4)(9 + 7n^4)
B Not a difference of squares because −49n^4 is not a perfect square.
C (9 − 7n^2)(9 − 7n^2)
D (9 + 7n^2 )(9 − 7n^2 )

Answer 1

We have

$81-49n^4=9^2-(7n^2)^2$

so this is indeed a difference of squares. Factorizing would give

$81-49n^4=(9-7n^2)(9+7n^2)$

making the answer D.

We can further factor the first term here and write

$9-7n^2=3^2-(\sqrt 7\,n)^2=(3-\sqrt7\,n)(3+\sqrt7\,n)$

but that's clearly out of the scope of this question.