Question

How do you factor `81x^4-1`?

Answer 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)`

We can use this a couple of times to find:

`81x^4-1 = (9x^2)-1^2`

`color(white)(81x^4-1) = (9x^2-1)(9x^2+1)`

`color(white)(81x^4-1) = ((3x)^2-1^2)(9x^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`, so has no linear factors with real coefficients.

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

`9x^2+1 = (3x)^2-i^2 = (3x-i)(3x+i)`

So if we allow complex coefficients then:

`81x^4-1 = (3x-1)(3x+1)(3x-i)(3x+i)`