Solve for \( x \) \( -x^{3}-2 x^{2}+8 x=0 \) \( x=[?], \square, \square] \)

To solve the equation: \[ - x^{3} - 2x^{2} + 8x = 0 \] **Step 1: Factor out \( x \)** \[ x(-x^{2} - 2x + 8) = 0 \] This gives one solution: \[ x = 0 \] **Step 2: Solve the quadratic equation** \[ - x^{2} - 2x + 8 = 0 \] Multiply both sides by \(-1\) to simplify: \[ x^{2} + 2x - 8 = 0 \] **Step 3: Use the quadratic formula** \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] For the equation \(x^{2} + 2x - 8 = 0\), \(a = 1\), \(b = 2\), and \(c = -8\): \[ x = \frac{-2 \pm \sqrt{(2)^{2} - 4(1)(-8)}}{2(1)} \] \[ x = \frac{-2 \pm \sqrt{4 + 32}}{2} \] \[ x = \frac{-2 \pm \sqrt{36}}{2} \] \[ x = \frac{-2 \pm 6}{2} \] This yields two solutions: \[ x = \frac{-2 + 6}{2} = 2 \] \[ x = \frac{-2 - 6}{2} = -4 \] **Final Solutions:** \[ x = [0, \, 2, \, -4] \]