Final answer:
The polynomial function with integer coefficients and a leading coefficient of 1 that has x = sqrt(27) as a root must include the conjugate root x = -sqrt(27). Multiplying the corresponding factors together yields f(x) = x^2 - 27, which is not present among the given options, indicating a potential error in the provided choices.
Step-by-step explanation:
The question asks for a polynomial function f(x) with integer coefficients, a leading coefficient of 1, and x = sqrt(27) as one of its roots. Recognizing that the root x = sqrt(27) is an irrational number, we recall that polynomial equations with integer coefficients must have irrational roots in conjugate pairs. Therefore, if x = sqrt(27) is a root, then its conjugate x = -sqrt(27) must also be a root.
First, we express sqrt(27) in its simplest radical form, which is sqrt(27) = sqrt(9*3) = 3*sqrt(3). Consequently, the conjugate pair of roots are x = 3*sqrt(3) and x = -3*sqrt(3).
To find the polynomial, we will use these roots to create factors of the function. The factors corresponding to 3*sqrt(3) and -3*sqrt(3) are (x - 3*sqrt(3)) and (x + 3*sqrt(3)), respectively. When multiplied together, these factors yield a quadratic with integer coefficients:
(x - 3*sqrt(3)) * (x + 3*sqrt(3)) = x^2 - (3*sqrt(3))^2 = x^2 - 27
Thus, the simplest polynomial function satisfying the conditions of the question is f(x) = x^2 - 27, which none of the given answer options match. The given options all result in polynomials with roots at x = 3, x = -3, and either x = 1 or x = -1, none of which include x = sqrt(27) or its conjugate.
In conclusion, you would need to find a different polynomial function or possibly have a typo in the provided answer choices, as none of the options correspond to the requirement of having x = sqrt(27) as one of its roots.