Question

complex numbers online practice
complete this assessment to review what youve learned. it will not count toward your grade.
what must be the degree of a polynomial function with two x-intercepts and four imaginary roots? (1 point)
8
2
4
6

Explanation:

Step1: Recall Polynomial Root Rules

For a polynomial function, the number of real roots (x - intercepts) and imaginary roots (complex roots) relates to the degree. Real roots can be repeated, and complex roots come in conjugate pairs (for polynomials with real coefficients, which is assumed here). Each x - intercept is a real root. If there are two x - intercepts, these are real roots (they could be repeated, but here we consider their count for the degree). Imaginary roots: since they come in conjugate pairs, four imaginary roots mean \( \frac{4}{2}=2 \) pairs of conjugate complex roots. But actually, the number of roots (counting multiplicities) is equal to the degree. The total number of roots (real + imaginary, counting multiplicities) is the degree. Here, we have two real roots (x - intercepts) and four imaginary roots. So total roots: \( 2 + 4=6 \). So the degree of the polynomial is 6.

Step2: Verify with Polynomial Degree Concept

The degree of a polynomial is the highest power of x, and also equal to the number of roots (counting multiplicities) by the Fundamental Theorem of Algebra. So adding the number of real roots (2) and imaginary roots (4) gives the total number of roots, which is the degree. So \( 2 + 4=6 \).

Answer:

6 (corresponding to the option with 6, likely the last option in the list as per the problem's options: the option with 6)