Given that a polynomial function [tex]\( f(x) \)[/tex] has rational coefficients, and it has roots [tex]\( 0 \)[/tex], [tex]\( 4 \)[/tex], and [tex]\( 3 + \sqrt{11} \)[/tex], we need to determine which additional root must also be present.
Let's consider the properties of polynomials with rational coefficients. If a polynomial has rational coefficients and a root involving an irrational number, then the conjugate of that root must also be a root. This is because the coefficients of the polynomial must remain rational.
Alright, we have identified the root [tex]\( 3 + \sqrt{11} \)[/tex].
The conjugate of [tex]\( 3 + \sqrt{11} \)[/tex] is [tex]\( 3 - \sqrt{11} \)[/tex].
Therefore, [tex]\( 3 - \sqrt{11} \)[/tex] must also be a root of the polynomial function [tex]\( f(x) \)[/tex].
Among the provided options:
1. [tex]\( 3 + i \sqrt{11} \)[/tex]
2. [tex]\( -3 + i \sqrt{11} \)[/tex]
3. [tex]\( 3 - \sqrt{11} \)[/tex]
4. [tex]\( -3 - \sqrt{11} \)[/tex]
The correct answer that corresponds to the required conjugate is [tex]\( 3 - \sqrt{11} \)[/tex].
Hence, the additional root of [tex]\( f(x) \)[/tex] must be [tex]\( 3 - \sqrt{11} \)[/tex].
