Answer
$(-\infty,-2)U(-1,2)$. See graph.
Work Step by Step
Step 1. $x^3+x^2\lt 4x+4\longrightarrow x^2(x+1)\lt 4(x+1)\longrightarrow (x+1)(x^2-4)\lt0 \longrightarrow (x+1)(x+2)(x-2)\lt0$
Step 2. Identify boundary points $x=-2,-1,2$ and intervals $(-\infty,-2),(-2,-1),(-1,2),(2,\infty)$
Step 3. Choose test values $x=-3,-1.5,0,3$ for each interval and evaluate the inequality to get $True,\ False,\ True,\ False$
Step 4. Thus we have the solution interval(s) $(-\infty,-2)U(-1,2)$. See graph.
