Solve the inequality algebraically. \[ x^{3}-2 x^{2}-80 x<0 \]

To solve the inequality \( x^{3}-2 x^{2}-80 x<0 \), we first factor the left-hand side. 1. Factor out the common term \( x \): \[ x(x^{2} - 2x - 80) < 0 \] 2. Now, we need to factor the quadratic \( x^{2} - 2x - 80 \). We can look for two numbers that multiply to \(-80\) and add to \(-2\). These numbers are \(8\) and \(-10\). Thus, we can factor it as: \[ x(x - 10)(x + 8) < 0 \] 3. To find the values of \( x \) that satisfy this inequality, we identify the critical points by setting each factor to zero: \[ x = 0, \quad x = 10, \quad x = -8 \] 4. We will now test intervals around these critical points to see where the product \( x(x - 10)(x + 8) \) is negative. The critical points divide the real number line into the intervals: - \( (-\infty, -8) \) - \( (-8, 0) \) - \( (0, 10) \) - \( (10, \infty) \) 5. Choose test points from each interval: - For \( x = -9 \) in \( (-\infty, -8) \): \[ (-9)(-9 - 10)(-9 + 8) \rightarrow (-9)(-19)(-1) < 0 \quad \text{(negative)} \] - For \( x = -4 \) in \( (-8, 0) \): \[ (-4)(-4 - 10)(-4 + 8) \rightarrow (-4)(-14)(4) > 0 \quad \text{(positive)} \] - For \( x = 5 \) in \( (0, 10) \): \[ (5)(5 - 10)(5 + 8) \rightarrow (5)(-5)(13) < 0 \quad \text{(negative)} \] - For \( x = 11 \) in \( (10, \infty) \): \[ (11)(11 - 10)(11 + 8) \rightarrow (11)(1)(19) > 0 \quad \text{(positive)} \] 6. Combining the results, the solution where the product is negative occurs in the intervals: \[ (-\infty, -8) \quad \text{and} \quad (0, 10) \] 7. Therefore, the final solution to the inequality \( x^{3}-2 x^{2}-80 x<0 \) is: \[ \boxed{(-\infty, -8) \cup (0, 10)} \]