Answer
$$x = - 3$$
Work Step by Step
$$\eqalign{
& {x^3} + 3{x^2} + 4x + 12 = 0 \cr
& {\text{factoring by grouping terms}} \cr
& {x^2}\left( {x + 3} \right) + 4\left( {x + 3} \right) = 0 \cr
& \left( {x + 3} \right)\left( {{x^2} + 4} \right) = 0 \cr
& {\text{By the zero product property}} \cr
& x + 3 = 0{\text{ or }}{x^2} + 4 = 0 \cr
& {\text{Then the real solution is}} \cr
& x = - 3 \cr} $$
