When an $n^{\text{th}}$ polynomial have $n/2$ roots and the other $n/2$is their conjugates?

Question

First of all, i'm really new on complex analysis and i never took algebra course. So i'm very sorry if my question is looks so dumb. (And sorry for my bad English).

Sometimes, for example we meet a polynomial with 4 roots, 2 of them are basic numbers and the other 2 is their conjugates. And we sometimes meet polynomial with 2 roots which is has no imaginary root (all of them are real numbers) that mean this quadratic has no conjugate of its root.

Then, when i saw on youtube, This is the link : https://youtu.be/z1rjJ_hbbgs

He has sextic polynomial (a polynomial of degree six) in his case, the sextic polynomial equation is the following :

$$r^6=-1$$

With three roots of them are :

$$\cos{\left(\dfrac{\pi}{6}\right)}+i\sin{\left(\dfrac{\pi}{6}\right)}$$

$$\cos{\left(\dfrac{\pi}{2}\right)}+i\sin{\left(\dfrac{\pi}{2}\right)}$$

$$\cos{\left(\dfrac{5\pi}{6}\right)}+i\sin{\left(\dfrac{5\pi}{6}\right)}$$

He claims that we can find the other roots which are their conjugates.

Then i was thinking about "he has $n^{th}$ polynomial in this case $n=6$ and then $\frac n2$ which is $3$ roots of them are the other roots conjugates."

How do we know that? In what case we could use that?

I think, i need to add some preliminaries.

e.g 1

$(z-2)^{10}$ is a tenth degree polynomial which has $10$ similar real roots which is $2$. But he has no conjugates even the imaginary roots.

e.g 2

$z^8+256$ is an Octic polynomial which has $4$ roots which four of them are the conjugates of the other $4$.

e.g 3

$z^6-z^3+z=0$ have 2 distinct roots and their conjugates and 2 real roots for the other.

In short, my question is : In what case we can find the $n^{th}$ degree polynomial has $\frac n 2$ roots and the other $\frac n2$ is their conjugates?

Or...

(another question) How do we know that the solutions have complex numbers and their conjugates (even if it might not be $\frac n2$)

I hope my question is clear. And ask me below if it's difficult to understand.

Answer 1

If a polynomial with real coefficients does not have any real root then first of all its degree must be even and it does not change sign.

For example if all coefficients are positive or all are negative.

If the absolute minimum is positive or the absolute maximum is negative then all roots are complex conjugates

Answer 2

For any polynomial equation $ \sum\limits_{k=0}^{n} a_kx^{k}$ with real coefficients, the conjugate of any root is also a root. You can prove this by simply taking complex conjugate in the equation $ \sum\limits_{k=0}^{n} a_kx^{k}=0$. If the equation is of even degree you can pair roots as stated.

Answer 3

If a polynomial has two conjugate roots, let $x\pm iy$, it must be a multiple of

$$(z-x-iy)(z-x+iy)=z^2-2xz+x^2+y^2.$$

Notice that all coefficients are real. In fact, the roots of a polynomial of real coefficients are either reals or pairs of complex conjugates. Such a polynomial can always be factored as the product of quadratic and linear factors with real coefficients.