Question

A polynomial P has degree 6 and has 2 real roots. According to the Fundamental Theorem of Algebra, how many non-real complex roots must P have?

(A) 0
(B) 2
(C) 4
(D) 6

Answer 1

According to the Fundamental Theorem of Algebra, a 6th degree polynomial must have exactly 6 roots. If it has 2 real roots, then there must be 4 non-real complex roots to fulfill the requirement of 6 total roots. The correct answer is (C) 4.

Step-by-step explanation:

The student's question involves the application of the Fundamental Theorem of Algebra, which states that every non-constant single-variable polynomial with complex coefficients has as many roots as its degree, counted with multiplicity. For a polynomial of degree 6, this means it must have exactly 6 roots.

If the polynomial P has 2 real roots, then the remaining roots must be complex. We know that non-real complex roots of polynomials with real coefficients always come in conjugate pairs. Therefore, if there are 2 real roots, there must be 6 - 2 = 4 non-real complex roots.

The correct answer to the question is: (C) 4. These are the four non-real complex roots that, along with the 2 real roots, make up the total of 6 roots required by the Fundamental Theorem of Algebra for a 6th degree polynomial.