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

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

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Answer 1 · 2017-02-11T22:19:25

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

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

Hence we find:

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

$16 x^{4} - 81 y^{4} = (4 x^{2} - 9 y^{2}) (4 x^{2} + 9 y^{2})$ $color(white)(16x^4-81y^4) = (4x^2-9y^2)(4x^2+9y^2)$

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

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

The remaining quadratic factor has no linear factors with Real coefficients.

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

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Explanation:

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