Score: \( 0 / 4 \) Penalty: 1 off Question Solve the equation for all values of \( x \) \[ 3 x(8 x-9)\left(x^{2}+64\right)=0 \]

To solve the equation \( 3 x(8 x - 9)(x^2 + 64) = 0 \), we need to set each factor equal to zero. 1. First factor: \( 3x = 0 \) - This gives us \( x = 0 \). 2. Second factor: \( 8x - 9 = 0 \) - Solving this, we get: \[ 8x = 9 \\ x = \frac{9}{8} \] 3. Third factor: \( x^2 + 64 = 0 \) - Rearranging gives us: \[ x^2 = -64 \] - Since the square of a real number cannot be negative, this does not yield any real solutions. Instead, we find complex solutions: \[ x = \pm 8i \] Combining all the solutions, we have: \[ x = 0, \quad x = \frac{9}{8}, \quad x = 8i, \quad x = -8i \] So, the complete solution set is: \[ \{0, \frac{9}{8}, 8i, -8i\} \]