Simplify \( \frac{2}{z+2}+\frac{4}{5 z}+\frac{4}{z^{2}+2 z} \)

To simplify the expression \[ \frac{2}{z+2}+\frac{4}{5z}+\frac{4}{z^{2}+2z}, \] we start by factoring the last term: \[ z^2 + 2z = z(z + 2). \] Thus, we rewrite the expression as: \[ \frac{2}{z+2} + \frac{4}{5z} + \frac{4}{z(z+2)}. \] Now, we can combine these fractions. The least common denominator (LCD) is \(5z(z + 2)\). We'll convert each fraction to have this common denominator: 1. For \(\frac{2}{z+2}\): \[ \frac{2}{z+2} \cdot \frac{5z}{5z} = \frac{10z}{5z(z+2)}. \] 2. For \(\frac{4}{5z}\): \[ \frac{4}{5z} \cdot \frac{(z+2)}{(z+2)} = \frac{4(z+2)}{5z(z+2)} = \frac{4z + 8}{5z(z+2)}. \] 3. For \(\frac{4}{z(z+2)}\): \[ \frac{4}{z(z+2)} \cdot \frac{5}{5} = \frac{20}{5z(z+2)}. \] Now we can combine these fractions: \[ \frac{10z + 4z + 8 + 20}{5z(z + 2)} = \frac{14z + 28}{5z(z + 2)}. \] Next, we can factor the numerator: \[ 14z + 28 = 14(z + 2). \] Thus, we have: \[ \frac{14(z + 2)}{5z(z + 2)}. \] Now, we can cancel \(z + 2\) from the numerator and the denominator (assuming \(z \neq -2\)): \[ \frac{14}{5z}. \] Therefore, the simplified form is: \[ \frac{14}{5z}. \]