If a polynomial function, [tex]f(x)[/tex], with rational coefficients has roots 0, 4, and [tex]3+\sqrt{11}[/tex], what must also be a root of [tex]f(x)[/tex]?

A. [tex]3-\sqrt{11}[/tex]
B. [tex]-3+i\sqrt{11}[/tex]
C. [tex]3-\sqrt{11}[/tex]
D. [tex]-3-\sqrt{11}[/tex]



Answer :

To solve this problem, we need to understand the property of polynomials with rational coefficients and how their roots behave, especially when they include irrational numbers like [tex]\( \sqrt{11} \)[/tex].

Step 1: Identify the given roots - We know that the given roots of the polynomial are:
- [tex]\( 0 \)[/tex]
- [tex]\( 4 \)[/tex]
- [tex]\( 3 + \sqrt{11} \)[/tex]

Step 2: Recall the principle - For polynomials with rational coefficients, if a non-rational root [tex]\( a + b\sqrt{c} \)[/tex] exists, its conjugate [tex]\( a - b\sqrt{c} \)[/tex] must also be a root to ensure the polynomial coefficients remain rational. This is because any irrational parts need to cancel out when forming the polynomial equation.

Step 3: Identify the conjugate root - The conjugate of the root [tex]\( 3 + \sqrt{11} \)[/tex] is [tex]\( 3 - \sqrt{11} \)[/tex].

Step 4: Conclusion - Therefore, the root [tex]\( 3 - \sqrt{11} \)[/tex] must also be a root of the polynomial function [tex]\( f(x) \)[/tex].

So, the correct answer is:
[tex]\[ 3 - \sqrt{11} \][/tex]

This matches our choice:

[tex]\[ \boxed{3 - \sqrt{11}} \][/tex]