1. **State the problem:** We need to create a sign chart for the function $$g(x) = -x^4 + 2x^3 + 8x^2$$ to determine where the function is positive, negative, or zero. 2. **Find the roots of the function:** Set $$g(x) = 0$$ and solve for $$x$$. $$-x^4 + 2x^3 + 8x^2 = 0$$ 3. **Factor the expression:** $$-x^2(x^2 - 2x - 8) = 0$$ 4. **Solve each factor:** - From $$-x^2 = 0$$, we get $$x = 0$$. - From $$x^2 - 2x - 8 = 0$$, use the quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$$ So, $$x = \frac{2 + 6}{2} = 4$$ and $$x = \frac{2 - 6}{2} = -2$$. 5. **Roots are:** $$x = -2, 0, 4$$. 6. **Determine the sign of $$g(x)$$ in intervals divided by the roots:** Intervals: $$(-\infty, -2), (-2, 0), (0, 4), (4, \infty)$$. 7. **Test points in each interval:** - For $$x = -3$$ in $$(-\infty, -2)$$: $$g(-3) = -(-3)^4 + 2(-3)^3 + 8(-3)^2 = -81 - 54 + 72 = -63 < 0$$ - For $$x = -1$$ in $$(-2, 0)$$: $$g(-1) = -1 + (-2) + 8 = 5 > 0$$ - For $$x = 1$$ in $$(0, 4)$$: $$g(1) = -1 + 2 + 8 = 9 > 0$$ - For $$x = 5$$ in $$(4, \infty)$$: $$g(5) = -625 + 250 + 200 = -175 < 0$$ 8. **Sign chart summary:** - $$g(x) < 0$$ on $$(-\infty, -2) \cup (4, \infty)$$ - $$g(x) > 0$$ on $$(-2, 0) \cup (0, 4)$$ - $$g(x) = 0$$ at $$x = -2, 0, 4$$ **Final answer:** $$g(x)$$ is negative for $$x < -2$$ and $$x > 4$$, positive between $$-2$$ and $$4$$ except at $$0$$ where it is zero.