1. **Problem:** Find the zeros of the polynomial function $f(x) = 2x^3 - 3x^2 - 8x + 12$ using the Rational Root Theorem. 2. **Rational Root Theorem:** Possible rational roots are of the form $\pm \frac{p}{q}$ where $p$ divides the constant term and $q$ divides the leading coefficient. 3. For $f(x) = 2x^3 - 3x^2 - 8x + 12$, constant term is 12 and leading coefficient is 2. 4. Factors of 12: $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$. 5. Factors of 2: $\pm1, \pm2$. 6. Possible rational roots: $\pm1, \pm\frac{1}{2}, \pm2, \pm3, \pm\frac{3}{2}, \pm4, \pm6, \pm12$. 7. Test $x=2$: $f(2) = 2(8) - 3(4) - 8(2) + 12 = 16 - 12 - 16 + 12 = 0$, so $x=2$ is a root. 8. Divide $f(x)$ by $(x-2)$ using synthetic division: $$\begin{array}{r|rrrr} 2 & 2 & -3 & -8 & 12 \\ & & 4 & 2 & -12 \\ \hline & 2 & 1 & -6 & 0 \end{array}$$ 9. Quotient polynomial: $2x^2 + x - 6$. 10. Solve $2x^2 + x - 6 = 0$ using quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2 \times 2} = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm 7}{4}$$ 11. Roots from quadratic: $x = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2}$ and $x = \frac{-1 - 7}{4} = \frac{-8}{4} = -2$. 12. **Final zeros:** $x = 2, \frac{3}{2}, -2$. 1. **Problem:** Find the zeros of $g(x) = x^3 - 6x^2 + 11x - 6$ using the Rational Root Theorem. 2. Constant term: 6, leading coefficient: 1. 3. Possible rational roots: $\pm1, \pm2, \pm3, \pm6$. 4. Test $x=1$: $1 - 6 + 11 - 6 = 0$, root found. 5. Divide $g(x)$ by $(x-1)$: $$\begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array}$$ 6. Quotient: $x^2 - 5x + 6$. 7. Factor $x^2 - 5x + 6 = (x-2)(x-3)$. 8. **Final zeros:** $x = 1, 2, 3$. 1. **Problem:** Find zeros of $h(x) = 3x^3 + 2x^2 - 7x - 6$ using Rational Root Theorem. 2. Constant term: -6, leading coefficient: 3. 3. Factors of -6: $\pm1, \pm2, \pm3, \pm6$. 4. Factors of 3: $\pm1, \pm3$. 5. Possible rational roots: $\pm1, \pm\frac{1}{3}, \pm2, \pm\frac{2}{3}, \pm3, \pm6$. 6. Test $x=1$: $3 + 2 - 7 - 6 = -8 \neq 0$. 7. Test $x=2$: $3(8) + 2(4) - 7(2) - 6 = 24 + 8 - 14 - 6 = 12 \neq 0$. 8. Test $x=-1$: $-3 + 2 + 7 - 6 = 0$, root found. 9. Divide $h(x)$ by $(x+1)$: $$\begin{array}{r|rrrr} -1 & 3 & 2 & -7 & -6 \\ & & -3 & 1 & 6 \\ \hline & 3 & -1 & -6 & 0 \end{array}$$ 10. Quotient: $3x^2 - x - 6$. 11. Solve $3x^2 - x - 6 = 0$: $$x = \frac{1 \pm \sqrt{(-1)^2 - 4(3)(-6)}}{2 \times 3} = \frac{1 \pm \sqrt{1 + 72}}{6} = \frac{1 \pm \sqrt{73}}{6}$$ 12. **Final zeros:** $x = -1, \frac{1 + \sqrt{73}}{6}, \frac{1 - \sqrt{73}}{6}$.