Proving the Absence of Positive Real Roots for Polynomials

When analyzing the behavior of polynomials, particularly determining the presence of positive real roots, several mathematical techniques can be applied. This article explores how to prove that the polynomial f(x) x^6 - x^4 - x^2 - x 3 has no positive real roots. We will use both algebraic and graphical methods to arrive at this conclusion.

Step-by-Step Proof Using Algebraic Methods

Let's consider the polynomial f(x) x^6 - x^4 - x^2 - x 3. Our goal is to determine if this polynomial has any positive real roots. We will analyze the polynomial using algebraic techniques.

Step 1: Rewrite the Polynomial for Clarity

The given polynomial is:

Equation:

f(x) x^6 - x^4 - x^2 - x 3

Step 2: Evaluate f(x) at Specific Points

We will evaluate the polynomial at specific points to gain insights into its behavior.

At x 0 f(0) 0^6 - 0^4 - 0^2 - 0 3 3

At x 1 f(1) 1^6 - 1^4 - 1^2 - 1 3 1 - 1 - 1 - 1 3 1

Both values are positive, indicating that for these specific points, f(x) > 0.

Step 3: Check the Derivative for Monotonicity

We can find the derivative f'(x) to determine if the function is increasing for x > 0.

f'(x) 6x^5 - 4x^3 - 2x - 1

Let's analyze this derivative for x > 0.

For 6x^5, since x > 0, we have 6x^5 > 0.

For -4x^3, since x > 0, we have -4x^3 .

For -2x, since x > 0, we have -2x .

For -1, this is a constant negative value.

Summing these terms, we get that f'(x) > 0 for all x > 0. Therefore, the polynomial f(x) is strictly increasing on (0, infty).

Step 4: Conclude the Proof

Since f(0) 3 > 0 and f(x) is strictly increasing for x > 0, it follows that f(x) > 0 for all x > 0. Therefore, f(x) cannot have any positive real roots.

Final Statement:

Thus, we conclude that the polynomial x^6 - x^4 - x^2 - x 3 0 has no positive real roots.

Using Descartes' Rule of Signs for Verification

Another method to prove the absence of positive real roots is by using Descartes' Rule of Signs. For the polynomial:

f(x) x^6 - x^4 - x^2 - x 3

Count the number of sign changes in the sequence of coefficients: - - - . There is only one sign change. Therefore, according to Descartes' Rule of Signs, the polynomial can have at most one positive real root. Since we have already shown that f(x) > 0 for all x > 0, it follows that there are no positive real roots.

Beyond Positive Real Roots: Negative Real Roots

To further validate our proof, we can consider the possibility of negative real roots. For negative roots, we evaluate f(-x).

Equation:

f(-x) (-x)^6 - (-x)^4 - (-x)^2 - (-x) 3 x^6 - x^4 - x^2 x 3

By examining the signs, we see:

For x^6, - - 3, there are two sign changes.

According to Descartes' Rule of Signs, this polynomial can have at most two negative real roots.

Conclusion

In conclusion, we have proven that the polynomial x^6 - x^4 - x^2 - x 3 0 has no positive real roots by analyzing its behavior at specific points, evaluating its derivative, and applying Descartes' Rule of Signs. Additionally, the polynomial may have up to two negative real roots.