1. **State the problem:** We want to solve the inequality $$(x-3)(x^2+1)(x^2+2x-15) > 0.$$ 2. **Find the zeros (roots) of the polynomial:** The zeros occur where each factor equals zero. - For $x-3=0$, the root is $x=3$. - For $x^2+1=0$, there are no real roots since $x^2=-1$ has no real solution. - For $x^2+2x-15=0$, use the quadratic formula: $$x=\frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 60}}{2} = \frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2}.$$ This gives two roots: $$x=\frac{-2 - 8}{2} = -5, \quad x=\frac{-2 + 8}{2} = 3.$$ 3. **List all zeros in increasing order:** The roots are $-5$ and $3$. Note that $3$ appears twice because it is a root of both $x-3$ and $x^2+2x-15$. Therefore, the zeros are $$[-5, 3, 3].$$