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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]


Sagot :

To determine the required additional root for the polynomial function [tex]\( f(x) \)[/tex] with rational coefficients, let's start by understanding the properties of polynomials with rational coefficients.

1. Roots and Rational Coefficients:
- If a polynomial has rational coefficients and has an irrational root, the conjugate of that irrational root must also be a root of the polynomial.
- This means, for any root of the form [tex]\( a + \sqrt{b} \)[/tex], the root [tex]\( a - \sqrt{b} \)[/tex] must also be present if the polynomial is to have rational coefficients.

2. Given Roots:
- The polynomial [tex]\( f(x) \)[/tex] has the roots: [tex]\( 0 \)[/tex], [tex]\( 4 \)[/tex], and [tex]\( 3 + \sqrt{11} \)[/tex].

3. Finding the Conjugate Root:
- Since [tex]\( 3 + \sqrt{11} \)[/tex] is an irrational root, its conjugate [tex]\( 3 - \sqrt{11} \)[/tex] must also be a root to ensure the polynomial has rational coefficients.

4. Answer:
- Therefore, the root that must also be a root of [tex]\( f(x) \)[/tex] is [tex]\( 3 - \sqrt{11} \)[/tex].

Hence, the correct answer is [tex]\( 3 - \sqrt{11} \)[/tex].
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