1. **State the problem:** Find the zeros of the function $$f(x) = 2x^3 + 5x^2 - 11x + 4$$. 2. **Formula and rules:** To find zeros, solve $$f(x) = 0$$. For cubic polynomials, try rational root theorem candidates and factor. 3. **Apply rational root theorem:** Possible roots are factors of 4 over factors of 2: $$\pm1, \pm2, \pm4, \pm\frac{1}{2}, \pm\frac{2}{2}=\pm1, \pm\frac{4}{2}=\pm2$$. 4. **Test roots:** Evaluate $$f(1) = 2(1)^3 + 5(1)^2 - 11(1) + 4 = 2 + 5 - 11 + 4 = 0$$, so $$x=1$$ is a root. 5. **Divide polynomial by $$x-1$$:** Use synthetic division: Coefficients: 2 | 5 | -11 | 4 Bring down 2, multiply by 1: 2, add to 5: 7 Multiply 7 by 1: 7, add to -11: -4 Multiply -4 by 1: -4, add to 4: 0 Quotient: $$2x^2 + 7x - 4$$ 6. **Solve quadratic:** $$2x^2 + 7x - 4 = 0$$ Use quadratic formula: $$x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-4)}}{2(2)} = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm \sqrt{81}}{4}$$ 7. **Simplify roots:** $$\sqrt{81} = 9$$ So, $$x = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}$$ and $$x = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$$ 8. **Final answer:** The zeros of $$f(x)$$ are $$x = 1, \frac{1}{2}, -4$$.