Question 35827
This equation factors as the difference of two squares:
{{{x^4 - 16=0 }}}
{{{(x^2 - 4)(x^2 + 4) =0}}}


The first factor factors again:
{{{(x-2)(x+2)(x^2 + 4) = 0}}}


Now, set each factor equal to zero and solve:
{{{x=2}}} or {{{x= -2}}} or {{{x^2=-4}}}

{{{x=2}}} or {{{x= -2}}} or {{{x = sqrt(-4)}}} or {{{ x = - sqrt(-4) }}}
{{{x=2}}} or {{{x= -2}}} or {{{x= 2i}}} or {{{x= -2i }}}



R^2 at SCC