Question

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) one imaginary root two imaginary roots no imaginary roots three imaginary roots

Explanation:

Step1: Determine the degree of the polynomial

First, we rewrite the polynomial \( g(x) = 13x^3 - 3x^2 + 12x - 9 \) in standard form (descending powers of \( x \)). The degree of a polynomial is the highest power of \( x \) with a non - zero coefficient. For \( g(x) \), the highest power of \( x \) is 3, so the degree \( n = 3 \). By the Fundamental Theorem of Algebra, a polynomial of degree \( n \) has exactly \( n \) roots (counting multiplicities) in the complex number system.

Step2: Relate x - intercepts to real roots

An \( x \) - intercept of a function \( y = g(x) \) is a real root of the equation \( g(x)=0 \). We are given that \( g(x) \) has one \( x \) - intercept, which means it has one real root.

Step3: Calculate the number of imaginary roots

Let the number of real roots be \( r \) and the number of imaginary roots be \( i \). Since complex roots (imaginary roots are complex roots with non - zero imaginary parts) occur in conjugate pairs (for polynomials with real coefficients, which this polynomial has since all coefficients \( 13,-3,12, - 9 \) are real), the number of imaginary roots \( i \) must be even? Wait, no. Wait, the total number of roots (real + imaginary, and remember that imaginary roots come in pairs for polynomials with real coefficients) is equal to the degree. Wait, the degree is 3. The number of real roots \( r = 1 \). Let the number of imaginary roots be \( i \). But since imaginary roots come in pairs (because if \( a+bi \) is a root, then \( a - bi \) is also a root for a polynomial with real coefficients), the possible number of imaginary roots for a cubic (degree 3) polynomial: the total number of roots is 3. If there is 1 real root, then the number of imaginary roots must be \( 3 - 1=2 \) (because the two imaginary roots form a conjugate pair).

Answer:

two imaginary roots (the option: two imaginary roots)