Answer
The solution is $(-\infty,-5]\cup[2,\infty)$
The graph is:
Work Step by Step
$(x+3)^{2}(x-2)(x+5)\ge0$
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 $(-\infty,-5]\cup[2,\infty)$
