How do you factor $81 x^{4} - 1$ $81x^4-1$ ?

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How do you factor $81 x^{4} - 1$ $81x^4-1$ ?

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Answer 1 · 2018-04-27T18:08:59

$81 x^{4} - 1 = (3 x - 1) (3 x + 1) (9 x^{2} + 1)$ $81x^4-1 = (3x-1)(3x+1)(9x^2+1)$

Explanation:

The difference of squares identity can be written:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2 = (A-B)(A+B)$

We can use this a couple of times to find:

$81 x^{4} - 1 = (9 x^{2}) - 1^{2}$ $81x^4-1 = (9x^2)-1^2$

$81 x^{4} - 1 = (9 x^{2} - 1) (9 x^{2} + 1)$ $color(white)(81x^4-1) = (9x^2-1)(9x^2+1)$

$81 x^{4} - 1 = ({(3 x)}^{2} - 1^{2}) (9 x^{2} + 1)$ $color(white)(81x^4-1) = ((3x)^2-1^2)(9x^2+1)$

$81 x^{4} - 1 = (3 x - 1) (3 x + 1) (9 x^{2} + 1)$ $color(white)(81x^4-1) = (3x-1)(3x+1)(9x^2+1)$

The remaining quadratic factor is always positive for real values of $x$ $x$ , so has no linear factors with real coefficients.

We can treat it as a difference of squares using the imaginary unit $i$ $i$ , which satisfies $i^{2} = - 1$ $i^2=-1$ as follows:

$9 x^{2} + 1 = {(3 x)}^{2} - i^{2} = (3 x - i) (3 x + i)$ $9x^2+1 = (3x)^2-i^2 = (3x-i)(3x+i)$

So if we allow complex coefficients then:

$81 x^{4} - 1 = (3 x - 1) (3 x + 1) (3 x - i) (3 x + i)$ $81x^4-1 = (3x-1)(3x+1)(3x-i)(3x+i)$

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How do you factor $81 x^{4} - 1$ $81x^4-1$ ?

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