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According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function?

[tex]\pi(x) = \left(x^3 - 3x + 1\right)^2[/tex]

A. 2 roots
B. 3 roots
C. 6 roots
D. 9 roots


Sagot :

Certainly! Let's solve this step-by-step.

Given polynomial: [tex]\(\pi(x) = \left(x^3 - 3x + 1\right)^2\)[/tex].

1. Identify the inner polynomial:
The inner polynomial is [tex]\(x^3 - 3x + 1\)[/tex].

2. Determine the degree of the inner polynomial:
The degree of the polynomial [tex]\(x^3 - 3x + 1\)[/tex] is 3 because the highest power of [tex]\(x\)[/tex] in this polynomial is 3.

3. Consider the effect of squaring the polynomial:
The given polynomial is [tex]\(\left(x^3 - 3x + 1\right)^2\)[/tex]. When we square a polynomial, the degree of the polynomial is multiplied by 2.

Thus, the degree of [tex]\(\pi(x) = \left(x^3 - 3x + 1\right)^2\)[/tex] is [tex]\(3 \times 2 = 6\)[/tex].

4. Apply the Fundamental Theorem of Algebra:
The Fundamental Theorem of Algebra states that a polynomial of degree [tex]\(n\)[/tex] will have exactly [tex]\(n\)[/tex] roots, including their multiplicities.

Therefore, the number of roots for the polynomial function [tex]\(\pi(x) = \left(x^3 - 3x + 1\right)^2\)[/tex] is 6.

So, the correct answer is:

[tex]\[ 6 \text{ roots} \][/tex]