Answer
$\left( -\infty, -2] \cup \right[2,3]$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$x^3-3x^2-4x+12\leq 0$,
$x^2(x-3)-4(x-3) \leq0$,
$(x^2-4)(x-3)\leq 0$,
$(x-2)(x+2)(x-3) \leq 0,$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(x-2)(x+2)(x-3) =0$
$x=-2$ or $x=2$ or $x=3$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& &(a-2)(a+2)(a-3) & \\
(-\infty, -2) & -5 & (-)(-)(-)=(-) & T\\
(-2, 2) & 0 & (-)(+)(-)=(+) & F\\
(2,3) & \frac{3}{2} & (+)(+)(-)=(-)& T\\
(2,\infty) & 5 & (+)(+)(+)=(+) & F
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $\left( -\infty, -2] \cup \right[2,3]$
