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Apply the Fundamental Theorem of Algebra to determine how many imaginary roots g(x)=12x−3x2+13x3−9 will have, if you know it has one x-intercept.(1 point)%0D%0AResponses%0D%0A%0D%0Ano imaginary roots%0D%0Ano imaginary roots%0D%0A%0D%0Atwo imaginary roots%0D%0Atwo imaginary roots%0D%0A%0D%0Athree imaginary roots%0D%0Athree imaginary roots%0D%0A%0D%0Aone imaginary root%0D%0Aone imaginary root

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The Fundamental Theorem of Algebra states that a polynomial of degree \( n \) has exactly \( n \) roots (including real and complex roots, and counting multiplicities).

The function \( g(x) = 12x - 3x^2 + 13x^3 - 9 \) is a polynomial of degree 3 (the term with the highest power is \( 13x^3 \)). This means that it has exactly 3 roots in total.

You mentioned that the polynomial has one x-intercept. An x-intercept corresponds to a real root. Since a polynomial can have real roots and complex roots, we can infer how many of each type can be present based on the number of total roots:

  1. If there is one real root (the x-intercept), and since coefficients of real polynomials are real, the remaining roots must be complex and come in conjugate pairs. Therefore, if there is one real root, the other two roots must be complex conjugates.

Thus, with one real root, the polynomial will have:

  • 1 real root
  • 2 complex roots (which will be imaginary, since they are non-real roots)

Consequently, the answer to how many imaginary roots the polynomial has is: two imaginary roots.