Answer
$(-\infty,-5]\cup[-2,2]$.
Work Step by Step
1. $x^3+5x^2\le 4x+20 \Longrightarrow x^2(x+5)\le 4(x+5) \Longrightarrow x^2(x+5)-4(x+5)\le0 \Longrightarrow(x+5)(x+2)(x-2)\le0 $
2. Identify boundary points $x=-5,-2,2$ and form intervals $(-\infty,-5],[-5,-2],[-2,2],[2,\infty)$.
3. Choose test values for each interval $x=-6,-3,0,3$ and test the inequality to get $True,\ False,\ True,\ False$.
4. Thus we have the solution $(-\infty,-5]\cup[-2,2]$.
