1. **State the problem:** Find the possible rational zeros of the polynomial $$f(x) = x^4 + 4x^3 - x^2 - 16x - 12$$ and test them to find actual zeros. 2. **Use the Rational Root Theorem:** Possible rational zeros are of the form $$\pm \frac{p}{q}$$ where $p$ divides the constant term and $q$ divides the leading coefficient. 3. **Identify factors:** - Leading coefficient is 1, factors: $\pm 1$ - Constant term is $-12$, factors: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ 4. **Possible rational zeros:** $$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$$ 5. **Test zeros by substitution or synthetic division:** - Test $x=1$: $1 + 4 - 1 - 16 - 12 = -24 \neq 0$ - Test $x=-1$: $1 - 4 - 1 + 16 - 12 = 0$ (zero found) 6. **Divide polynomial by $(x+1)$:** $$\frac{x^4 + 4x^3 - x^2 - 16x - 12}{x+1} = x^3 + 3x^2 - 4x - 12$$ 7. **Repeat Rational Root Theorem on quotient:** Possible zeros: same as before. 8. **Test $x=2$:** $$2^3 + 3(2)^2 - 4(2) - 12 = 8 + 12 - 8 - 12 = 0$$ (zero found) 9. **Divide quotient by $(x-2)$:** $$\frac{x^3 + 3x^2 - 4x - 12}{x-2} = x^2 + 5x + 6$$ 10. **Factor quadratic:** $$x^2 + 5x + 6 = (x+2)(x+3)$$ 11. **Final factorization:** $$f(x) = (x+1)(x-2)(x+2)(x+3)$$ 12. **Zeros:** $$x = -1, 2, -2, -3$$ **Answer:** The rational zeros of $f(x)$ are $-3, -2, -1,$ and $2$.