Can a 5 Degree Polynomial Equation Have Only Two Real Roots and the Remaining Complex Roots?

A 5-degree polynomial equation can indeed have only two real roots and the remaining roots as complex. This scenario arises due to the fundamental properties of polynomial equations, specifically the nature of their roots and the constraints imposed by real coefficients.

Understanding Polynomial Roots

For any polynomial equation with real coefficients, the complex roots always appear in conjugate pairs. This means that if a complex number (a bi) (where (a) and (b) are real, and (i) is the imaginary unit) is a root, then its conjugate (a - bi) is also a root. This property is crucial in determining the possible combinations of roots for a polynomial.

The Specific Case of a 5-degree Polynomial

Let's consider a general 5-degree polynomial:

[f(x) a_5x^5 a_4x^4 a_3x^3 a_2x^2 a_1x a_0]

Since the polynomial has real coefficients, it is guaranteed that any complex roots will occur in conjugate pairs. Therefore, with 5 roots in total (counting multiplicities), the possible combinations are:

5 real roots. 3 real roots and 2 complex conjugate pairs (4 complex roots). 1 real root and 4 complex conjugate pairs (2 complex roots). 2 real roots and 3 complex roots (with real roots and complex pairs).

From this list, you can see that a 5-degree polynomial can indeed have 2 real roots and 3 complex roots, provided the complex roots come in conjugate pairs. For example, if one complex root is (a bi), then another root must be (a - bi), and the third complex root could be another pair, or the same complex number repeated.

Examples of Polynomials with 2 Real and 3 Complex Roots

Let's look at some specific examples:

Example 1

[y x^2 - 1x - x^{2n}]

This can be broken down into:

The first factor (x^2 - 1) has two complex roots: (x pm i). The second factor (x - 1) has one real root: (x 1). The third factor (x^{2n}) can be ignored as it does not contribute real or complex roots unless (x 0) is the only real root, which is a special case.

Note that (x 0) is not a complex root, but a real one. If we consider (x^{2n}) as a normalized case, then we can say the polynomial has 2 real roots (1 and 0) and 3 complex roots ((pm i)).

Example 2

[x^5 - x^2 0]

This can be factored as:

[x^2(x-1)(x^2 1) 0]

The roots are:

(x 0) (1 real root, twice) (x 1) (1 real root) (x^2 1 0 Rightarrow x pm i) (2 complex roots, conjugate pair)

Here, the polynomial has 3 real roots if we include multiplicities (2 from 0 and 1), and 2 complex roots (1 conjugate pair).

Example 3

[x^5 - 4x^4 6x^3 - 4x^2 x 0]

This can be factored as:

[x(x-1)^4 0]

The roots are:

(x 0) (1 real root, twice) (x 1) (1 real root, four times)

Here, the polynomial has only 3 real roots if we include multiplicities, and no complex roots.

Conclusion

In conclusion, a 5-degree polynomial can indeed have exactly two real roots and three complex roots, provided the complex roots come in conjugate pairs. This is possible due to the fundamental properties of polynomial equations with real coefficients. Understanding these properties can help in solving and analyzing complex polynomial equations.