1. The problem asks us to find a polynomial function with zeros at $-3$, $1$, and $2$. 2. Recall that if a polynomial has zeros at $r_1$, $r_2$, and $r_3$, then it can be written as: $$f(x) = a(x - r_1)(x - r_2)(x - r_3)$$ where $a$ is a nonzero constant (usually taken as 1 for the simplest polynomial). 3. Substitute the given zeros $-3$, $1$, and $2$: $$f(x) = (x - (-3))(x - 1)(x - 2) = (x + 3)(x - 1)(x - 2)$$ 4. First, multiply the first two factors: $$(x + 3)(x - 1) = x^2 - x + 3x - 3 = x^2 + 2x - 3$$ 5. Now multiply this result by the third factor $(x - 2)$: $$ (x^2 + 2x - 3)(x - 2) = x^3 - 2x^2 + 2x^2 - 4x - 3x + 6 = x^3 - 7x + 6$$ 6. Simplify the expression: $$f(x) = x^3 - 7x + 6$$ 7. Therefore, the polynomial function with zeros at $-3$, $1$, and $2$ is: $$\boxed{f(x) = x^3 - 7x + 6}$$