1. **State the problem:** Solve the polynomial equation $$x^4 + 4x^3 - 3x^2 - 16x - 4 = 0$$ given that one rational root is $$x = -2$$. 2. **Use synthetic division to divide the polynomial by $$x + 2$$:** Set up synthetic division with root $$-2$$: Coefficients: 1 (for $$x^4$$), 4 (for $$x^3$$), -3 (for $$x^2$$), -16 (for $$x$$), -4 (constant). Carry down 1. Multiply $$1 \times (-2) = -2$$, add to 4: $$4 + (-2) = 2$$. Multiply $$2 \times (-2) = -4$$, add to -3: $$-3 + (-4) = -7$$. Multiply $$-7 \times (-2) = 14$$, add to -16: $$-16 + 14 = -2$$. Multiply $$-2 \times (-2) = 4$$, add to -4: $$-4 + 4 = 0$$. Remainder is 0, confirming $$x = -2$$ is a root. The quotient polynomial is: $$x^3 + 2x^2 - 7x - 2$$ 3. **Solve the cubic equation $$x^3 + 2x^2 - 7x - 2 = 0$$:** Try rational roots using factors of constant term (-2): $$\pm1, \pm2$$. Test $$x=1$$: $$1^3 + 2(1)^2 - 7(1) - 2 = 1 + 2 - 7 - 2 = -6 \neq 0$$. Test $$x=-1$$: $$(-1)^3 + 2(-1)^2 - 7(-1) - 2 = -1 + 2 + 7 - 2 = 6 \neq 0$$. Test $$x=2$$: $$2^3 + 2(2)^2 - 7(2) - 2 = 8 + 8 - 14 - 2 = 0$$. So, $$x=2$$ is a root. 4. **Use synthetic division to divide $$x^3 + 2x^2 - 7x - 2$$ by $$x - 2$$:** Coefficients: 1, 2, -7, -2. Carry down 1. Multiply $$1 \times 2 = 2$$, add to 2: $$2 + 2 = 4$$. Multiply $$4 \times 2 = 8$$, add to -7: $$-7 + 8 = 1$$. Multiply $$1 \times 2 = 2$$, add to -2: $$-2 + 2 = 0$$. Remainder 0 confirms root. Quotient polynomial: $$x^2 + 4x + 1$$ 5. **Solve quadratic $$x^2 + 4x + 1 = 0$$ using quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}$$ 6. **Final solution set:** $$\boxed{\{-2, 2, -2 + \sqrt{3}, -2 - \sqrt{3}\}}$$