autumncoloredleaves
contestada

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 )

Respuesta :

LammettHash

We have

[tex]81-49n^4=9^2-(7n^2)^2[/tex]

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

[tex]81-49n^4=(9-7n^2)(9+7n^2)[/tex]

making the answer D.

We can further factor the first term here and write

[tex]9-7n^2=3^2-(\sqrt 7\,n)^2=(3-\sqrt7\,n)(3+\sqrt7\,n)[/tex]

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

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