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anemone
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Prove that the polynomial $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$ has 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.
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.
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.
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.
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.