Polynomial zeros of the 5th and 6th degree?

Question

There are as many zeros as there are degrees. Correct?
Since complex zeros come in pairs there are 3 distinct real zeros in 5th degree Polynomial. Correct?

What about a polynomial of 6 degrees?
I assume there must be at least 2 complex zeros and as many 4 distinct real zeros. Correct?

I am teaching myself this math without the help of a teacher and want to know if I am understanding what is going on.

Answer 1

Since complex zeros come in pairs there are 3 distinct real zeros in 5th degree Polynomial. Correct?

Not necessarily. Assuming a 5th degree polynomial with real coefficients, there will be a total of 5 zeros; an odd number of real zeros, 1, 3, or 5, and an even number of complex zeros in conjugate pairs, 0, 2, or 4 (0, 1, or 2 conjugate pairs)



Assuming a 6 degree polynomial with real coefficients, there will be a total of 6 zeros, of which there will be an even number of real zeros, 0, 2, 4, or 6, and an even number of complex zeros in conjugate pairs, 0, 2, 4, or 6 (0, 1, 2, or 3 conjugate pairs)



Note that a polynomial of any degree with real coefficients may, or may not have complex zeros. But if there are any complex zeros, they will occur in conjugate pairs so there is an even number of them. There will then be enough real zeros to make the total number of zeros equal to the degree of the polynomial.



Remember this is only true for polynomials with real coefficients. Polynomials with complex coefficients obey different rules than those for real coefficients.

Answer 2

It is true that there are as many zeros as the degree of a polynomial. However, the next few parts are not true.

Complex zeros do come in pairs (specifically conjugates), but only if there are some possibly. There will always be an even number of complex solutions.

In a fifth degree polynomial, there could be equations with either 1, 3 or 5 real zeros, since the number of complex zeros always has to be divisible by 2. There may also be repeated solution(s) where one zero counts as two of them (which doesn't mean that all real zeros are distinct). This applies to any polynomial with a degree that is odd.

In a sixth degree polynomial, there would be equations with either 0, 2, 4, or 6 real zeros, for the same reason as above. Again, there may also be repeated solution(s). This applies to any polynomial with a degree that is even.

As you see, the exact number of complex, and real, zeros cannot be determined with just the degree of the polynomial. You need to know the equation to know the 'possible' number of real zeros and complex zeros. You can only determine the number of zeros (complex or real).

There is a method to know the 'possible' number of real and complex zeros, also both positive and negative. It is called the "Descartes' Rule of Signs." Seeing an equation of any degree, the number of possible positive and negative zeros can be found and the number of complex zeros can be predicted from there.

The Descartes' Rule of Signs says that a polynomial equation has the number of positive zeros as the number of sign (positive and negative) changes in descending order of each non-zero coefficient of the equation. The number of negative zeros is the number of sign changes in descending order of each non-zero coefficient of the equation when replacing -x for x in the equation (remember that the negative x will make its coefficients their negative if power is odd, and will keep their coefficients the same if power is even). You don't stop there; after knowing that specific number of zeros, you will have to subtract that number continually by 2 until you reach 1 or 0. Those numbers will be the number of possible positive and negative numbers respectively. After knowing the number of real zeros, the number of complex zeros can be predicted.

Find the possible numbers of positive, negative, and complex zeros of the following equation:

x^5 - 5x^4 + 3x^2 - 4x + 2 = 0

There is a total of 5 zeros.

The number of positive real zeros is the number of sign changes in each coefficient of the equation.

From left to right, sign changes: +, -, +, -, +. There are a total of 4 positive zeros. Wait. Subtract by 2 until you reach 0. Those will be the possible numbers of positive zeros: 4, 2, or 0.

The number of negative real zeros is the number of sign changes in each coefficient of the equation when x is -x. In the equation, the -x substitution will be:

-x^5 - 5x^4 + 3x^2 + 4x + 2 = 0

The sign changes: -, -, +, +, +. There is 1 negative zero. Since this is 1, it does not need to be subtracted by 2. It is certain that there is 1 negative solution.

To find the number of complex zeros, make possible arrangements of zeros with positive and negative.

1. 4 positive, 1 negative (which means 0 complex)

2. 2 positive, 1 negative (which means 2 complex)

3. 0 positive, 1 negative (which means 4 complex)

There is either 0, 2, or 4 complex zeros.

You have found each possible number of positive, negative, and complex zeros. In order to know how many real zeros there are, you have to either look at a graph, or do calculations.

Remember that there isn't a way to know the number of real (positive and negative) or complex zeros, especially when just given a degree. You can only predict the number possible for each of them when given a polynomial equation.

Answer 3

6th Degree Polynomial

Answer 4

TomV gave you a complete answer. Since you are teaching yourself, I thought some additonal info might be helpful. You can find all this online or in books.



I'm sure you know the quadratic equation that gives you the roots of a 2nd degree polynomial. The solution may of course involve square roots.



There are techniques, although quite messy, for finding the exact solution of 3rd and 4th degree polynomials. Because they are so messy, no one ever uses them, and they are rarely studied, although centuries ago they were important issues, and solutions were closely guarded secrets because they were so difficult. We say that 3rd and 4th degree polynomials are also "solvable by radicals" since the exact solutions involve square roots, cube roots, and 4th roots. It's far easier, and usually of more practical use, to just find a numeric solution using Newton's method or something similar.



However there is no general solution to 5th and higher degree polynomials. The solutions to polynomials with rational or integer coefficients are called "algebraic numbers", but they are simply not expressible in terms of radicals. There are exceptions for simple cases, like x^5 -2 has a root of 2^(1/5), but the general case does not have solutions like this.

Answer 5

For x = one million/2, multiply the two aspects through 2 2x = one million 2x - one million = 0 So while x = one million/2, (2x - one million) is a element with integer coefficients: f(x) = (x - 5) (x + 5) (x - 6) (2x - one million) f(x) = (x² - 25) (2x² - 13x + 6) f(x) = 2x? - 13x³ - 44x² + 325 x - one hundred fifty or you may only have prolonged your comprehensive polynomial through 2