1. **State the problem:** Find the polynomial function with zeros at $-3$, $1$, and $2$. 2. **Recall the fact:** If a polynomial has zeros at $a$, $b$, and $c$, then the polynomial can be written as $$f(x) = k(x - a)(x - b)(x - c)$$ where $k$ is a nonzero constant. 3. **Apply the zeros:** Here, the zeros are $-3$, $1$, and $2$, so the polynomial is $$f(x) = k(x - (-3))(x - 1)(x - 2) = k(x + 3)(x - 1)(x - 2)$$. 4. **Assume $k=1$ for the simplest polynomial:** $$f(x) = (x + 3)(x - 1)(x - 2)$$. 5. **Expand step-by-step:** First, multiply the last two factors: $$ (x - 1)(x - 2) = x^2 - 2x - x + 2 = x^2 - 3x + 2 $$ Then multiply by $(x + 3)$: $$ (x + 3)(x^2 - 3x + 2) = x(x^2 - 3x + 2) + 3(x^2 - 3x + 2) $$ $$ = x^3 - 3x^2 + 2x + 3x^2 - 9x + 6 $$ 6. **Simplify:** $$ x^3 - 3x^2 + 2x + 3x^2 - 9x + 6 = x^3 + ( -3x^2 + 3x^2 ) + (2x - 9x) + 6 = x^3 - 7x + 6 $$ 7. **Final polynomial function:** $$ \boxed{f(x) = x^3 - 7x + 6}$$ This polynomial has zeros at $-3$, $1$, and $2$ as required.