Determine whether 81-49n^4 is a difference of 2 squares. If so, factor it, if not explain why.?

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Answer 1 · 2018-06-12T15:04:10

Please read below;

$81 - 49 n^{4}$ $81 - 49n^4$

Is not a difference of $2$ $2$ squares because..

Difference of two squares is denoted by: $x^{2} - y^{2}$ $x^2 - y^2$

Hence;

$81 - 49 n^{4}$ $81 - 49n^4$ can be simplified as $= 9^{2} - 7^{2} n^{4}$ $= 9^2 - 7^2n^4$

Assuming it was;

$9^{2} - 7^{2} n^{2}$ $9^2 - 7^2n^2$ then it would have been difference of two squares $= (9 + 7 n) (9 - 7 n)$ $=(9 + 7n) (9- 7n)$