1. **State the problem:** Construct a polynomial function $f(x)$ of degree 4 with zeros at $-1$ (multiplicity 1), $2$ (multiplicity 2), and $3$ (multiplicity 1), and that passes through the point $(1, -12)$. 2. **Write the general form using zeros and multiplicities:** Since zeros are $-1$, $2$ (with multiplicity 2), and $3$, the factored form is: $$f(x) = a(x + 1)(x - 2)^2(x - 3)$$ where $a$ is a constant to be determined. 3. **Use the point $(1, -12)$ to find $a$:** Substitute $x=1$ and $f(1) = -12$: $$-12 = a(1 + 1)(1 - 2)^2(1 - 3)$$ Calculate each factor: $$-12 = a(2)(-1)^2(-2) = a(2)(1)(-2) = a(-4)$$ 4. **Solve for $a$:** $$a = \frac{-12}{-4} = 3$$ 5. **Write the final polynomial function:** $$f(x) = 3(x + 1)(x - 2)^2(x - 3)$$ This polynomial has degree 4, the specified zeros with correct multiplicities, and passes through the point $(1, -12)$.