Question 89996
Find the real zeros of:
{{{p(x) = x(x+1)(x^2-81)}}} 
To find the zeros, just set p(x) = 0 and, since the function is already factored, apply the zero products principle:
{{{x(x+1)(x^2-81) = 0}}} therefore:
{{{x = 0}}} or {{{x+1 = 0}}} or {{{x^2-81 = 0}}}
The roots are:
{{{x = 0}}}
{{{x = -1}}}
{{{x^2 = 81}}} so {{{x = 9}}} or {{{x = -9}}}