X(x-2)/x+3<0

To solve the inequality \(\frac{x(x-2)}{x+3} < 0\), we will follow these steps: 1. **Find critical points**: The critical points can be found by determining where the numerator and denominator are equal to zero. - **Numerator**: \(x(x-2) = 0\) - Solutions are \(x = 0\) and \(x = 2\). - **Denominator**: \(x + 3 = 0\) - Solution is \(x = -3\). So the critical points are \(x = -3\), \(x = 0\), and \(x = 2\). 2. **Create intervals**: The critical points divide the number line into the following intervals: - \((- \infty, -3)\) - \((-3, 0)\) - \((0, 2)\) - \((2, \infty)\) 3. **Test each interval**: We will choose a test point from each interval to determine if the expression is negative in that interval. - **Interval \((- \infty, -3)\)**: Choose \(x = -4\) \[ \frac{-4(-4-2)}{-4+3} = \frac{-4(-6)}{-1} = \frac{24}{-1} = -24 < 0 \] - **Interval \((-3, 0)\)**: Choose \(x = -1\) \[ \frac{-1(-1-2)}{-1+3} = \frac{-1(-3)}{2} = \frac{3}{2} > 0 \] - **Interval \((0, 2)\)**: Choose \(x = 1\) \[ \frac{1(1-2)}{1+3} = \frac{1(-1)}{4} = \frac{-1}{4} < 0 \] - **Interval \((2, \infty)\)**: Choose \(x = 3\) \[ \frac{3(3-2)}{3+3} = \frac{3(1)}{6} = \frac{3}{6} = \frac{1}{2} > 0 \] 4. **Conclusion from test points**: The expression is negative in the intervals \((- \infty, -3)\) and \((0, 2)\). 5. **Include/exclude critical points**: - At \(x = -3\), the denominator is zero, so it is not included. - At \(x = 0\) and \(x = 2\), the entire expression equals zero, which does not satisfy the strict inequality. 6. **Final solution**: The solution to the inequality \(\frac{x(x-2)}{x+3} < 0\) is: \[ \boxed{(-\infty, -3) \cup (0, 2)} \]