# How do you find all solutions to #x^4-81=0#?

##### 1 Answer

Aug 12, 2016

The zeros of this quartic are:

#### Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Hence we find:

#0 = x^4-81#

#= (x^2)^2-9^2#

#= (x^2-9)(x^2+9)#

#= (x^2-3^2)(x^2-(3i)^2)#

#= (x-3)(x+3)(x-3i)(x+3i)#

Hence zeros: