Question

For the equation x^7 - 2x^5 - 4x^3 - 2x - 1 = 0, state the number of complex roots, the possible number of real roots, and the possible rational roots.

A) 7 complex roots, 0 real roots, 0 rational roots
B) 7 complex roots, 0 real roots, 7 rational roots
C) 4 complex roots, 3 real roots, 4 rational roots
D) 4 complex roots, 3 real roots, 0 rational roots

Answer 1

The equation has 7 complex roots, 3 possible real roots, and no rational roots.

Step-by-step explanation:

The given equation is x^7 - 2x^5 - 4x^3 - 2x - 1 = 0.

To determine the number of complex roots, we can use the Fundamental Theorem of Algebra. According to the theorem, a polynomial equation of degree n has n complex roots. In this case, the equation is of degree 7, so it has 7 complex roots.

For the possible number of real roots, we can use the Descartes' rule of signs. By counting the sign changes in the coefficients, we find that there are at most 3 positive real roots and no negative real roots.

As for the possible rational roots, we can use the Rational Root Theorem. The theorem states that any rational root of the equation must be of the form p/q, where p is a factor of the constant term (in this case -1) and q is a factor of the leading coefficient (in this case 1).

By checking all possible combinations, we find that there are no rational roots.