1. **Problem statement:** Find a polynomial $f(x)$ with roots 0, 1, and -1 having multiplicities 3, 2, and 1 respectively, and such that $f(2) = 24$. 2. **General form:** A polynomial with roots $r_i$ and multiplicities $m_i$ can be written as $$f(x) = a(x - r_1)^{m_1}(x - r_2)^{m_2} \cdots$$ where $a$ is a constant. 3. **Apply roots and multiplicities:** Here, $$f(x) = a x^3 (x - 1)^2 (x + 1)^1 = a x^3 (x - 1)^2 (x + 1)$$ 4. **Use the condition $f(2) = 24$ to find $a$:** Calculate $f(2)$: $$f(2) = a \cdot 2^3 \cdot (2 - 1)^2 \cdot (2 + 1) = a \cdot 8 \cdot 1^2 \cdot 3 = 24a$$ Set equal to 24: $$24a = 24 \implies a = 1$$ 5. **Final polynomial:** $$f(x) = x^3 (x - 1)^2 (x + 1)$$ 6. **Optional expansion:** First expand $(x - 1)^2 = x^2 - 2x + 1$, then multiply by $(x + 1)$: $$(x^2 - 2x + 1)(x + 1) = x^3 - x^2 - x + 1$$ Finally multiply by $x^3$: $$f(x) = x^3 (x^3 - x^2 - x + 1) = x^6 - x^5 - x^4 + x^3$$ **Answer:** $$\boxed{f(x) = x^3 (x - 1)^2 (x + 1)}$$