Solve the following rational inequality \( \frac{x-2}{x^{2}-16}>0 \) State your answer using interval notation. Use \( U \) for union and oo for \( \infty \)

To solve the rational inequality \( \frac{x-2}{x^{2}-16}>0 \), we first factor the denominator: 1. **Factor the Denominator**: \[ x^2 - 16 = (x-4)(x+4) \] 2. **Sign Analysis**: The inequality can be rewritten as: \[ \frac{x-2}{(x-4)(x+4)} > 0 \] The critical points are where the numerator and denominator are zero: \(x = 2\), \( x = 4\), and \( x = -4\). These points divide the number line into intervals: \( (-\infty, -4) \), \( (-4, 2) \), \( (2, 4) \), and \( (4, \infty) \). 3. **Test each interval**: - **Interval \( (-\infty, -4) \)**: Choose \( x = -5 \): \[ \frac{-5-2}{(-5-4)(-5+4)} = \frac{-7}{(-9)(-1)} = \frac{-7}{9} < 0 \] - **Interval \( (-4, 2) \)**: Choose \( x = 0 \): \[ \frac{0-2}{(0-4)(0+4)} = \frac{-2}{(-4)(4)} = \frac{-2}{-16} = \frac{1}{8} > 0 \] - **Interval \( (2, 4) \)**: Choose \( x = 3 \): \[ \frac{3-2}{(3-4)(3+4)} = \frac{1}{(-1)(7)} = \frac{1}{-7} < 0 \] - **Interval \( (4, \infty) \)**: Choose \( x = 5 \): \[ \frac{5-2}{(5-4)(5+4)} = \frac{3}{(1)(9)} = \frac{3}{9} > 0 \] 4. **Constructing the Solution**: The intervals where the fraction is positive are: \((-4, 2)\) and \((4, \infty)\). 5. **Check endpoints**: The inequality does not include points where the rational expression is zero or undefined. Thus, we exclude \(x = -4\), \(x = 2\), and \(x = 4\). The final solution in interval notation is: \[ (-4, 2) U (4, oo) \]