Solve the inequality below. Give the solution using interval notation. Give all values as integers or reduced fractions. (Do not convert fractions to decimals.) \( \frac{\mathrm{x}^{2}-2 x}{\mathrm{x}+4}>0 \)

To solve the inequality \( \frac{x^{2}-2x}{x+4}>0 \), we first need to factor the numerator. The numerator \(x^{2} - 2x\) can be factored as \(x(x - 2)\). Thus, we rewrite the inequality as: \[ \frac{x(x - 2)}{x + 4} > 0 \] Next, we find the critical points by setting the numerator and denominator to zero: 1. \(x(x - 2) = 0\) gives \(x = 0\) and \(x = 2\). 2. \(x + 4 = 0\) gives \(x = -4\). Now, we have the critical points: \(x = -4\), \(x = 0\), and \(x = 2\). These points divide the number line into intervals. We need to test each interval to determine where the inequality holds true: 1. For the interval \((- \infty, -4)\), choose \(x = -5\): \[ \frac{-5(-5 - 2)}{-5 + 4} = \frac{-5 \cdot -7}{-1} = \frac{35}{-1} < 0 \quad (\text{not included}) \] 2. For the interval \((-4, 0)\), choose \(x = -1\): \[ \frac{-1(-1 - 2)}{-1 + 4} = \frac{-1 \cdot -3}{3} = \frac{3}{3} > 0 \quad (\text{included}) \] 3. For the interval \((0, 2)\), choose \(x = 1\): \[ \frac{1(1 - 2)}{1 + 4} = \frac{1 \cdot -1}{5} = \frac{-1}{5} < 0 \quad (\text{not included}) \] 4. For the interval \((2, \infty)\), choose \(x = 3\): \[ \frac{3(3 - 2)}{3 + 4} = \frac{3 \cdot 1}{7} > 0 \quad (\text{included}) \] Now, we combine the intervals where the inequality is satisfied. The critical points where the expression is zero or undefined are \(x = -4\) (undefined), \(x = 0\) (zero), and \(x = 2\) (zero). Thus, the solution in interval notation is: \[ (-4, 0) \cup (2, \infty) \]