1. **Problem statement:** Given the family of functions defined for $k > 0$ by $$f_k(x) = \frac{1}{2k} \cdot x^2 \cdot (x - 2k)^2,$$ show that for each $k$ there are exactly two zeros and find them. 2. **Finding zeros:** To find zeros, solve $f_k(x) = 0$: $$\frac{1}{2k} \cdot x^2 \cdot (x - 2k)^2 = 0.$$ Since $\frac{1}{2k} \neq 0$ for $k > 0$, zeros occur when either $$x^2 = 0 \quad \text{or} \quad (x - 2k)^2 = 0.$$ This gives $$x = 0 \quad \text{or} \quad x = 2k.$$ 3. **Number of zeros:** Both zeros are distinct for $k > 0$ because $0 \neq 2k$. Each zero has multiplicity 2 (due to the squares), so the graph touches the x-axis at these points but does not cross it. 4. **Conclusion:** For each $k > 0$, $f_k$ has exactly two zeros at $$x = 0 \quad \text{and} \quad x = 2k.$$