1. **State the problem:** Find all zeros of the polynomial function $$f(x) = x^4 + 2x^3 - 13x^2 - 14x + 24$$. 2. **Recall the method:** To find zeros of a polynomial, we try to factor it or use the Rational Root Theorem to test possible roots. 3. **Apply Rational Root Theorem:** Possible rational roots are factors of 24 divided by factors of 1, i.e., $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$$. 4. **Test roots:** Evaluate $f(x)$ at these values. - $f(1) = 1 + 2 - 13 - 14 + 24 = 0$ so $x=1$ is a root. - $f(-2) = 16 - 16 - 52 + 28 + 24 = 0$ so $x=-2$ is a root. - $f(3) = 81 + 54 - 117 - 42 + 24 = 0$ so $x=3$ is a root. - $f(-4) = 256 - 128 - 208 + 56 + 24 = 0$ so $x=-4$ is a root. 5. **Factor polynomial:** Since $x=1, -2, 3, -4$ are roots, the polynomial factors as $$f(x) = (x - 1)(x + 2)(x - 3)(x + 4)$$ 6. **List zeros:** The zeros are $$x = -4, -2, 1, 3$$. 7. **Check for multiplicity:** Each root appears once, so no double roots. **Final answer:** $$x = -4, -2, 1, 3$$