Apply the Fundamental Theorem of Algebra to find the number of imaginary roots for the polynomial f(x)=4x2−14+x8%0D%0A𝑓%0D%0A(%0D%0A𝑥%0D%0A)%0D%0A=%0D%0A4%0D%0A𝑥%0D%0A2%0D%0A−%0D%0A14%0D%0A+%0D%0A𝑥%0D%0A8%0D%0A, if you know its graph has two x-intercepts.(1 point)

Answer 1

To apply the Fundamental Theorem of Algebra to the polynomial \( f(x) = 4x^2 - 14 + x^8 \), we first need to determine the degree of the polynomial.

1. Identify the degree of the polynomial: The highest power of \( x \) present in the polynomial \( f(x) \) is \( x^8 \). Therefore, the degree of the polynomial is 8.

2. According to the Fundamental Theorem of Algebra, a polynomial of degree \( n \) has exactly \( n \) roots (counting multiplicities) in the complex number system.

Since the degree of \( f(x) \) is 8, it has 8 total roots.

3. We are given that the graph of \( f(x) \) has two \( x \)-intercepts. The \( x \)-intercepts correspond to the real roots of the polynomial.

From the information given:

• Real roots: 2 (corresponding to the 2 \( x \)-intercepts).
• Since complex roots appear in conjugate pairs, the remaining roots must account for the degree of the polynomial.

4. Now calculate the remaining roots:
   • Total roots = 8
   • Real roots = 2
   • Thus, the remaining roots = \( 8 - 2 = 6 \)

Since the polynomial degree is even (8), the remaining 6 roots must be complex, and because they can't all be real (given that we only have 2 real roots), we conclude that the remaining roots must be 3 pairs of imaginary roots.

5. Therefore, the number of imaginary (non-real) roots in \( f(x) \) is 6.