To determine another root of the polynomial function [tex]\( f(x) \)[/tex] with rational coefficients, given the roots 3 and [tex]\( \sqrt{7} \)[/tex], we need to first understand a crucial property of polynomials with rational coefficients. Specifically, if a polynomial has rational coefficients and includes any irrational roots, those roots must occur in conjugate pairs.
Given the roots 3 and [tex]\( \sqrt{7} \)[/tex]:
1. Identify the irrational root:
- Since [tex]\( \sqrt{7} \)[/tex] is an irrational number, for the polynomial to have rational coefficients, its conjugate, [tex]\( -\sqrt{7} \)[/tex], must also be a root of the polynomial.
2. Confirm the additional root:
- With 3 being a rational number, it does not necessitate a conjugate pair. Therefore, the only missing pair to complete the set of roots with rational coefficients is [tex]\( -\sqrt{7} \)[/tex].
Hence, another root of the polynomial must be [tex]\( -\sqrt{7} \)[/tex].
Therefore, the correct answer is:
[tex]\[ -\sqrt{7} \][/tex]
This is aligned with the calculated numerical result:
[tex]\[ -2.6457513110645907 \][/tex]
So, the additional root that must be a part of the polynomial [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{-\sqrt{7}} \][/tex]
