1. **Problem statement:** Find the leading coefficient $a$ of a fifth-degree polynomial with rational coefficients given zeros at $-3$, $4$ (multiplicity 2), and $1 + \sqrt{5}$, and the value $f(1) = 60$. 2. **Write the polynomial in factored form:** Since $1 + \sqrt{5}$ is a root and coefficients are rational, its conjugate $1 - \sqrt{5}$ is also a root. The quadratic factor from these roots is: $$x^2 - 2x - 3$$ The polynomial is: $$f(x) = a (x + 3)(x - 4)^2 (x^2 - 2x - 3)$$ 3. **Use the given value $f(1) = 60$ to find $a$:** Substitute $x=1$: $$60 = a (1 + 3)(1 - 4)^2 (1^2 - 2 \cdot 1 - 3)$$ Calculate each term: $$(1 + 3) = 4$$ $$(1 - 4)^2 = (-3)^2 = 9$$ $$1^2 - 2 \cdot 1 - 3 = 1 - 2 - 3 = -4$$ So: $$60 = a \times 4 \times 9 \times (-4) = a \times (-144)$$ 4. **Solve for $a$:** $$a = \frac{60}{-144} = -\frac{5}{12}$$ 5. **Final polynomial:** $$f(x) = -\frac{5}{12} (x + 3)(x - 4)^2 (x^2 - 2x - 3)$$ **Answer:** The leading coefficient is $\boxed{-\frac{5}{12}}$.