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

Aug 12, 2016

The zeros of this quartic are: $3 , - 3 , 3 i , - 3 i$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Hence we find:

$0 = {x}^{4} - 81$

$= {\left({x}^{2}\right)}^{2} - {9}^{2}$

$= \left({x}^{2} - 9\right) \left({x}^{2} + 9\right)$

$= \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {\left(3 i\right)}^{2}\right)$

$= \left(x - 3\right) \left(x + 3\right) \left(x - 3 i\right) \left(x + 3 i\right)$

Hence zeros: $3 , - 3 , 3 i , - 3 i$