How do you factor $81 - 16 x^{4}$ $81-16x^4$ ?

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How do you factor $81 - 16 x^{4}$ $81-16x^4$ ?

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Answer 1 · 2015-11-12T07:06:30

Apply the difference of squares formula twice to find
$81 - 16 x^{4} = (9 + 4 x^{2}) (3 + 2 x) (3 - 2 x)$ $81-16x^4 = (9 + 4x^2)(3+2x)(3-2x)$

Explanation:

The difference of squares formula states
$x^{2} - y^{2} = (x + y) (x - y)$ $x^2 - y^2 = (x+y)(x-y)$

$81 = 9^{2}$ $81 = 9^2$ and $16 x^{4} = {(4 x^{2})}^{2}$ $16x^4 = (4x^2)^2$

So, applying the formula,
$81 - 16 x^{4} = 9^{2} - {(4 x^{2})}^{2} = (9 + 4 x^{2}) (9 - 4 x^{2})$ $81 - 16x^4 = 9^2 - (4x^2)^2 = (9 + 4x^2)(9 - 4x^2)$

But we are not quite done.

$9 = 3^{2}$ $9 = 3^2$ and $4 x^{2} = {(2 x)}^{2}$ $4x^2 = (2x)^2$

So applying the formula to $9 - 4 x^{2}$ $9-4x^2$ , we get the final result of

$81 - 16 x^{4} = (9 + 4 x^{2}) (9 - 4 x^{2}) = (9 + 4 x^{2}) (3 + 2 x) (3 - 2 x)$ $81 - 16x^4 = (9 + 4x^2)(9 - 4x^2) = (9 + 4x^2)(3+2x)(3-2x)$

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How do you factor $81 - 16 x^{4}$ $81-16x^4$ ?

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