1. **State the problem:** Find all real zeros of the function $$g(x) = 2x^3 + x^2 - 40x - 20$$ using synthetic division and rational zeros to determine irrational zeros. 2. **Identify possible rational zeros:** By the Rational Root Theorem, possible rational zeros are factors of the constant term $-20$ divided by factors of the leading coefficient $2$. These are $$\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20, \pm \frac{1}{2}, \pm \frac{5}{2}$$. 3. **Test rational zeros using synthetic division:** Start with $x=2$. Synthetic division setup: $$\begin{array}{r|rrrr} 2 & 2 & 1 & -40 & -20 \\ & & 4 & 10 & -60 \\ \hline & 2 & 5 & -30 & -80 \end{array}$$ Remainder is $-80 \neq 0$, so $x=2$ is not a zero. Try $x=-2$: $$\begin{array}{r|rrrr} -2 & 2 & 1 & -40 & -20 \\ & & -4 & 6 & 68 \\ \hline & 2 & -3 & -34 & 48 \end{array}$$ Remainder $48 \neq 0$, so $x=-2$ is not a zero. Try $x=5$: $$\begin{array}{r|rrrr} 5 & 2 & 1 & -40 & -20 \\ & & 10 & 55 & 75 \\ \hline & 2 & 11 & 15 & 55 \end{array}$$ Remainder $55 \neq 0$, so $x=5$ is not a zero. Try $x=-1$: $$\begin{array}{r|rrrr} -1 & 2 & 1 & -40 & -20 \\ & & -2 & 1 & 39 \\ \hline & 2 & -1 & -39 & 19 \end{array}$$ Remainder $19 \neq 0$, so $x=-1$ is not a zero. Try $x=\frac{1}{2}$: $$\begin{array}{r|rrrr} \frac{1}{2} & 2 & 1 & -40 & -20 \\ & & 1 & 1 & -19.5 \\ \hline & 2 & 2 & -39 & -39.5 \end{array}$$ Remainder $-39.5 \neq 0$, so $x=\frac{1}{2}$ is not a zero. Try $x=-5$: $$\begin{array}{r|rrrr} -5 & 2 & 1 & -40 & -20 \\ & & -10 & 45 & -25 \\ \hline & 2 & -9 & 5 & -45 \end{array}$$ Remainder $-45 \neq 0$, so $x=-5$ is not a zero. Try $x=4$: $$\begin{array}{r|rrrr} 4 & 2 & 1 & -40 & -20 \\ & & 8 & 36 & -16 \\ \hline & 2 & 9 & -4 & -36 \end{array}$$ Remainder $-36 \neq 0$, so $x=4$ is not a zero. Try $x=-\frac{1}{2}$: $$\begin{array}{r|rrrr} -\frac{1}{2} & 2 & 1 & -40 & -20 \\ & & -1 & 0 & 20 \\ \hline & 2 & 0 & -40 & 0 \end{array}$$ Remainder is $0$, so $x = -\frac{1}{2}$ is a zero. 4. **Divide polynomial by $(x + \frac{1}{2})$:** The quotient is $$2x^2 + 0x - 40 = 2x^2 - 40$$. 5. **Solve quadratic $2x^2 - 40 = 0$:** $$2x^2 - 40 = 0$$ $$2x^2 = 40$$ $$\cancel{2}x^2 = \cancel{2}20$$ $$x^2 = 20$$ $$x = \pm \sqrt{20} = \pm 2\sqrt{5}$$ 6. **Final answer:** The real zeros of $g(x)$ are $$\left\{ -\frac{1}{2}, -2\sqrt{5}, 2\sqrt{5} \right\}$$ These are listed least to greatest in reduced radical and fraction form.