No Real Zeroes of $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$

MHB
Thread starter anemone
Start date Apr 16, 2021

In summary, the degree of the polynomial is 6 and there are no real zeroes. The Descartes' Rule of Signs can be used to determine the number of possible real zeroes, and in this case, there are none. It is possible to find complex zeroes using the Fundamental Theorem of Algebra, but their exact values cannot be determined without advanced methods. Despite having no real zeroes, the polynomial can still be graphed by plotting points or using a graphing calculator. Lastly, the polynomial cannot be factored and is considered a prime polynomial, further supporting the absence of real zeroes.

Apr 16, 2021

anemone

Gold Member

MHB

POTW Director

Prove that the polynomial $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$ has no real zeroes.

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Jul 29, 2021

maxkor

Related to No Real Zeroes of $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$

1. What does it mean for a polynomial to have no real zeroes?

When a polynomial has no real zeroes, it means that there are no values of x that make the polynomial equal to zero when plugged in. In other words, the polynomial does not intersect or touch the x-axis at any point.

2. How can you determine if a polynomial has any real zeroes?

One way to determine if a polynomial has any real zeroes is by graphing it and looking for points where the graph intersects or touches the x-axis. Another way is by using the Rational Zero Theorem to find possible rational zeroes and then using synthetic division or the Remainder Theorem to test if they are actually zeroes.

3. Is it possible for a polynomial to have no real zeroes?

Yes, it is possible for a polynomial to have no real zeroes. For example, the polynomial in this question, $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$, has no real zeroes because its graph does not intersect or touch the x-axis at any point.

4. What is the degree of a polynomial with no real zeroes?

The degree of a polynomial with no real zeroes can vary. In the case of $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$, the degree is 6. However, a polynomial with no real zeroes can have any degree, as long as it does not intersect or touch the x-axis.

5. Can a polynomial with no real zeroes have complex zeroes?

Yes, a polynomial with no real zeroes can have complex zeroes. Complex zeroes are solutions that involve the imaginary number i, which is defined as the square root of -1. In the case of $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$, the polynomial has no real zeroes but it has complex zeroes.