Answer
$\text{the interval } [2,\infty)$
Work Step by Step
Expressing the given inequality in factored form results to
\begin{array}{l}
x^3+2x^2-4x\ge8
\\\\
x^3+2x^2-4x-8\ge0
\\\\
(x^3+2x^2)-(4x+8)\ge0
\\\\
x^2(x+2)-4(x+2)\ge0
\\\\
(x+2)(x^2-4)\ge0
\\\\
(x+2)(x+2)(x-2)\ge0
.\end{array}
Hence, the critical points are $x=\{ -2,2 \}$.
If $x\lt-2$, then,
\begin{array}{l}
(x+2)(x+2)(x-2)
\\\Rightarrow
(-)(-)(-)
\\=\text{ negative}
.\end{array}
If $-2\lt x\lt2$, then,
\begin{array}{l}
(x+2)(x+2)(x-2)
\\\Rightarrow
(+)(+)(-)
\\=\text{ negative}
.\end{array}
If $x\gt2$, then,
\begin{array}{l}
(x+2)(x+2)(x-2)
\\\Rightarrow
(+)(+)(+)
\\=\text{ positive}
.\end{array}
Since the factored form of the inequality uses $\ge0$, then take the positive result. Hence, the solution set is $
\text{the interval } [2,\infty)
$.
