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

1 Answer
Apr 27, 2018

Answer:

#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)#