1. **State the problem:** We are given a polynomial function $f(x)$ such that $$f(x) f(1) = f(x) + f(1)$$ and the value $$f(3) = 28.$$ We need to find $$f(4).$$ 2. **Analyze the given equation:** The equation can be rewritten as $$f(x) f(1) = f(x) + f(1).$$ Rearranging, $$f(x) f(1) - f(x) = f(1)$$ $$f(x)(f(1) - 1) = f(1).$$ 3. **Consider cases:** - If $f(1) = 1$, then the left side becomes $f(x)(0) = 0$ and the right side is $1$, which is impossible. - Therefore, $f(1) \neq 1$ and we can divide both sides by $(f(1) - 1)$: $$f(x) = \frac{f(1)}{f(1) - 1}.$$ 4. **Interpretation:** This means $f(x)$ is a constant function for all $x$, equal to $$f(x) = c = \frac{f(1)}{f(1) - 1}.$$ 5. **Use the given value $f(3) = 28$:** Since $f$ is constant, $$c = 28.$$ 6. **Find $f(1)$:** From the constant value formula, $$28 = \frac{f(1)}{f(1) - 1}.$$ Multiply both sides by $(f(1) - 1)$: $$28(f(1) - 1) = f(1).$$ 7. **Simplify:** $$28 f(1) - 28 = f(1)$$ $$28 f(1) - f(1) = 28$$ $$(28 - 1) f(1) = 28$$ $$27 f(1) = 28$$ $$f(1) = \frac{28}{27}.$$ 8. **Find $f(4)$:** Since $f$ is constant, $$f(4) = 28.$$ **Final answer:** $$\boxed{28}.$$