1. **State the problem:** Find a polynomial function in factored form with zeros 2 (of multiplicity 2) and 3, passing through the point (1, 6). 2. **Write the general form:** Since zeros are 2 (order 2) and 3, the polynomial can be written as $$f(x) = a(x-2)^2(x-3)$$ where $a$ is a constant to be determined. 3. **Use the point to find $a$:** Substitute $x=1$ and $f(1)=6$ into the equation: $$6 = a(1-2)^2(1-3)$$ 4. **Calculate the values inside the parentheses:** $$(1-2)^2 = (-1)^2 = 1$$ $$(1-3) = -2$$ 5. **Substitute these values:** $$6 = a \times 1 \times (-2) = -2a$$ 6. **Solve for $a$:** $$a = \frac{6}{-2} = -3$$ 7. **Write the final factored form:** $$f(x) = -3(x-2)^2(x-3)$$ This is the polynomial function with the given zeros and passing through the point (1, 6).