How do you factor #81x^4-1#?
1 Answer
Apr 27, 2018
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
We can treat it as a difference of squares using the imaginary unit
#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)#