How do you use the difference of two squares formula to factor $x^{4} n - 81$ $x^4n - 81$ ?

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Answer 1 · 2015-05-03T12:43:42

I have to wonder if the $n$ $n$ in your expression is an error; but assuming it isn't:

The general equation for the difference of two squares is
$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2 = (a+b)(a-b)$

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

$= {(\sqrt{n} x^{2})}^{2} - 9^{2}$ $=(sqrt(n)x^2)^2 - 9^2$

$= (\sqrt{n} x^{2} + 9) (\sqrt{n} x^{2} - 9)$ $=(sqrt(n)x^2 +9)(sqrt(n)x^2-9)$

$= (\sqrt{n} x^{2} + 9) (\sqrt[4]{n} x + 3) (\sqrt[4]{n} x - 3)$ $=(sqrt(n)x^2+9)(root(4)(n)x+3)(root(4)(n)x-3)$

How do you use the difference of two squares formula to factor x^4n - 81 x4n−81x^4n - 81 ?