## Precalculus: Mathematics for Calculus, 7th Edition

The solution is $(-\infty,-5]\cup[2,\infty)$ The graph is:
$(x+3)^{2}(x-2)(x+5)\ge0$ Begin immediately by finding the intervals, because all nonzero terms are on one side of the inequality and the nonzero side is given in factored form. The factors are $(x+3)^{2}$ $,$ $x-2$ and $x+5$. Set them equal to $0$ and solve for $x$: $(x+3)^{2}=0$ $x+3=0$ $x=-3$ $x-2=0$ $x=2$ $x+5=0$ $x=-5$ The factors are zero when $x=-3,2,-5$. These three numbers divide the real line into the following intervals: $(-\infty,-5)$ $,$ $(-5,-3)$ $,$ $(-3,2)$ $,$ $(2,\infty)$ Elaborate a diagram, using test point to determine the sign of each factor in each interval: (refer to the attached image below) It can be seen from the diagram that the inequality is satisfied in the intervals $(-\infty,-5)$ and $(2,\infty)$. Also, the inequality involves $\ge$ so the endpoints also satisfy the inequality. The solution is $(-\infty,-5]\cup[2,\infty)$