1. **Problem statement:** Find all real zeros of the function $$f(x) = x^3 - 6x^2 + 11x - 6$$ and determine if any zeros have multiplicity greater than one. 2. **Formula and rules:** To find zeros of a polynomial, solve $$f(x) = 0$$. Zeros with multiplicity greater than one mean the factor appears more than once in the factorization. 3. **Step 1: Try to factor the cubic polynomial.** Use Rational Root Theorem to test possible roots among factors of constant term 6: $$\pm1, \pm2, \pm3, \pm6$$. 4. **Step 2: Test $x=1$:** $$f(1) = 1 - 6 + 11 - 6 = 0$$ so $x=1$ is a root. 5. **Step 3: Divide $f(x)$ by $(x-1)$ using polynomial division or synthetic division:** $$f(x) = (x-1)(x^2 - 5x + 6)$$ 6. **Step 4: Factor the quadratic:** $$x^2 - 5x + 6 = (x-2)(x-3)$$ 7. **Step 5: Complete factorization:** $$f(x) = (x-1)(x-2)(x-3)$$ 8. **Step 6: Identify zeros and multiplicities:** Zeros are $x=1, 2, 3$ each with multiplicity 1 (since each factor appears once). **Final answer:** The real zeros are $$x=1, x=2, x=3$$ and none have multiplicity greater than one.