Question 24416
You can apply the zero product principle to this problem: If{{{a*b = 0}}} then either {{{a = 0}}} or {{{b = 0}}} or both.
{{{x(x^2+4)(x^2-x-6) = 0}}} 
{{{x = 0}}} and/or {{{x^2+4 = 0}}} and/or {{{x^2-x-6 = 0}}}
One of the real roots is x = 0
If {{{x^2+4 = 0}}} then {{{x^2 = -4}}} and x = +or-{{{sqrt(-4)}}} so these two roots are:
x = 2i and x = -2i  These are not real roots
If {{{x^2-x-6 = 0}}} then {{{(x+2)(x-3) = 0}}} so {{{x = -2}}} and {{{x = 3}}} These two roots are real.

The real roots are:
x = 0
x = -2
x = 3