Answer
$(-\infty,-5)\cup(-5, -3)\cup(1, \infty)$
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.
$(x+5)^2(x+3)(x-1) \gt 0$
$f(x)=(x+5)^2(x+3)(x-1) $
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(x+5)^2(x+3)(x-1) =0$
$x=-5$ or $x=-3$ or $x=1$
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,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\
& & (a+5)^2(a+3)(a-1) & \\
(-\infty,-5) & -6 & (+)(-)(-) & T\\
(-5,-3) & -4 & (+)(-)(-) & T\\
(-3,1) & 0 & (+)(+)(-) & F\\
(1,\infty) & 2 & (+)(+)(+) & T
\end{array}$
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: $(-\infty,-5)\cup(-5, -3)\cup(1, \infty)$
