1. **State the problem:** Find all zeros of the cubic function $$f(x) = 2x^3 + x^2 - 12x + 9$$. These are the values of $x$ where $f(x) = 0$. 2. **Use the Rational Root Theorem:** Possible rational roots are factors of the constant term 9 divided by factors of the leading coefficient 2, i.e., $\pm1, \pm3, \pm9, \pm\frac{1}{2}, \pm\frac{3}{2}, \pm\frac{9}{2}$. 3. **Test possible roots:** Evaluate $f(x)$ at these values to find actual roots. - $f(1) = 2(1)^3 + 1^2 - 12(1) + 9 = 2 + 1 - 12 + 9 = 0$ so $x=1$ is a root. 4. **Divide the polynomial by $(x-1)$:** Use synthetic division or polynomial division. $$\begin{array}{r|rrrr} 1 & 2 & 1 & -12 & 9 \\ \hline & 2 & 3 & -9 & 0 \end{array}$$ The quotient is $2x^2 + 3x - 9$. 5. **Solve the quadratic $2x^2 + 3x - 9 = 0$ using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4(2)(-9)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 72}}{4} = \frac{-3 \pm \sqrt{81}}{4}$$ 6. **Simplify the square root:** $$\sqrt{81} = 9$$ 7. **Find the two roots:** $$x = \frac{-3 + 9}{4} = \frac{6}{4} = \frac{3}{2}$$ $$x = \frac{-3 - 9}{4} = \frac{-12}{4} = -3$$ 8. **List all roots from smallest to largest:** $$x = -3, 1, \frac{3}{2}$$ There are no repeated roots. **Final answer:** $$x = -3, 1, \frac{3}{2}$$