1. **State the problem:** Find all zeros of the polynomial function $$f(x) = 3x^4 + 11x^3 + 11x^2 + x - 2$$ and list them from smallest to largest, including multiplicities. 2. **Recall the Rational Root Theorem:** Possible rational roots are factors of the constant term divided by factors of the leading coefficient. Here, factors of $-2$ are $\pm1, \pm2$ and factors of $3$ are $\pm1, \pm3$. Possible rational roots: $\pm1, \pm2, \pm\frac{1}{3}, \pm\frac{2}{3}$. 3. **Test possible roots by substitution or synthetic division:** - Test $x=1$: $3(1)^4 + 11(1)^3 + 11(1)^2 + 1 - 2 = 3 + 11 + 11 + 1 - 2 = 24 \neq 0$ - Test $x=-1$: $3(-1)^4 + 11(-1)^3 + 11(-1)^2 + (-1) - 2 = 3 - 11 + 11 - 1 - 2 = 0$ so $x=-1$ is a root. 4. **Divide $f(x)$ by $(x+1)$ using synthetic division:** $$\begin{array}{r|rrrrr} -1 & 3 & 11 & 11 & 1 & -2 \\ & & -3 & -8 & -3 & 2 \\ \hline & 3 & 8 & 3 & -2 & 0 \end{array}$$ Quotient: $3x^3 + 8x^2 + 3x - 2$ 5. **Find roots of the cubic $3x^3 + 8x^2 + 3x - 2$:** Possible rational roots: $\pm1, \pm2, \pm\frac{1}{3}, \pm\frac{2}{3}$. - Test $x=\frac{1}{3}$: $$3\left(\frac{1}{3}\right)^3 + 8\left(\frac{1}{3}\right)^2 + 3\left(\frac{1}{3}\right) - 2 = 3\cdot\frac{1}{27} + 8\cdot\frac{1}{9} + 1 - 2 = \frac{1}{9} + \frac{8}{9} + 1 - 2 = 1 + 1 - 2 = 0$$ So $x=\frac{1}{3}$ is a root. 6. **Divide the cubic by $(x - \frac{1}{3})$:** Multiply divisor by 3 to avoid fractions: divide by $(3x - 1)$. Synthetic division: $$\begin{array}{r|rrrr} \frac{1}{3} & 3 & 8 & 3 & -2 \\ & & 1 & 3 & 2 \\ \hline & 3 & 9 & 6 & 0 \end{array}$$ Quotient: $3x^2 + 9x + 6$ 7. **Solve quadratic $3x^2 + 9x + 6 = 0$ using quadratic formula:** $$x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 3 \cdot 6}}{2 \cdot 3} = \frac{-9 \pm \sqrt{81 - 72}}{6} = \frac{-9 \pm \sqrt{9}}{6} = \frac{-9 \pm 3}{6}$$ 8. **Calculate roots:** - $$x = \frac{-9 + 3}{6} = \frac{-6}{6} = -1$$ - $$x = \frac{-9 - 3}{6} = \frac{-12}{6} = -2$$ 9. **List all roots including multiplicities:** - From step 3: $x = -1$ - From step 6 and 8: $x = \frac{1}{3}$, $x = -1$, $x = -2$ So roots are $-2$, $-1$ (twice), and $\frac{1}{3}$. **Final answer:** $$x = -2, -1, -1, \frac{1}{3}$$