Solve the following equation completely. There are 3 answerboxes listed. \( x^{3}+2 x^{2}-2 x-4=0 \)

To solve the equation \( x^{3}+2 x^{2}-2 x-4=0 \), let's first look for possible rational roots using the Rational Root Theorem. Testing small integers, we find that \( x = 1 \) is a root. Now we can perform synthetic division to divide the polynomial by \( x - 1 \): ``` 1 | 1 2 -2 -4 | 1 3 1 -------------------- 1 3 1 -3 ``` This gives us the quotient \( x^2 + 3x + 4 \) and a remainder of \( 0 \). Thus, we can factor our original polynomial as: \[ (x - 1)(x^2 + 3x + 4) = 0 \] Next, we need to solve the quadratic \( x^2 + 3x + 4 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1, b = 3, c = 4 \): \[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 16}}{2} = \frac{-3 \pm \sqrt{-7}}{2} \] This gives us complex roots: \[ x = \frac{-3}{2} \pm \frac{\sqrt{7}}{2}i \] Thus, the complete list of solutions for the equation \( x^{3}+2 x^{2}-2 x-4=0 \) is: 1. \( x = 1 \) 2. \( x = \frac{-3}{2} + \frac{\sqrt{7}}{2}i \) 3. \( x = \frac{-3}{2} - \frac{\sqrt{7}}{2}i \) So, the complete answer boxes will be: 1. \( 1 \) 2. \( \frac{-3}{2} + \frac{\sqrt{7}}{2}i \) 3. \( \frac{-3}{2} - \frac{\sqrt{7}}{2}i \)