jarriouswesley7
Answered

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

Sagot :

GinnyAnswer
When working with polynomial functions that have rational coefficients, an important property to consider is that any non-rational root must appear in conjugate pairs. This is necessary to ensure that the polynomial retains rational coefficients.

Given the roots of the polynomial function [tex]\( f(x) \)[/tex]:
1. [tex]\(0\)[/tex]
2. [tex]\(4\)[/tex]
3. [tex]\(3 + \sqrt{11}\)[/tex]

Let’s analyze the roots step by step:
- [tex]\(0\)[/tex] and [tex]\(4\)[/tex] are rational numbers, so they do not affect the rationality of the polynomial's coefficients.
- However, [tex]\(3 + \sqrt{11}\)[/tex] is not a rational number. Since the polynomial has rational coefficients, the root [tex]\( 3 + \sqrt{11} \)[/tex] must be paired with its conjugate to ensure that all coefficients remain rational.

The conjugate of [tex]\( 3 + \sqrt{11} \)[/tex] is [tex]\( 3 - \sqrt{11} \)[/tex].

Therefore, by the principle that non-rational roots must appear in conjugate pairs, the polynomial function [tex]\( f(x) \)[/tex] must also have the root [tex]\( 3 - \sqrt{11} \)[/tex].

Thus, the correct answer is:

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