I was watching the Khan Academy video on the Fundamental Theorem of Algebra when I got confused by something that Sal Khan states. From what I understand, the Theorem says that the complex zeros of a polynomial function always come in pairs because they are conjugates of each other. But exactly what would happen with a 3rd degree polynomial with no real zeros? Why can you not have 3 complex zeros with a 3rd degree polynomial but it is possible with a 4th degree polynomial?
Finally, what then would the zeros of a 3rd degree polynomial with no real zeros be?
This is the video:
https://www.khanacademy.org/math/algebra2/polynomial_and_rational/fundamental-theorem-of-algebra/v/fundamental-theorem-of-algebra-intro
Answer
Why can you not have $3$ complex zeros with a $3$rd degree polynomial, but it is possible with a $4$th degree polynomial?
Observation: You can, if the coefficients are themselves complex.
Hint: Odd exponents maintain the sign, and even ones suppress it. Notice the obvious difference between the graphics of cubic and quartic $($or even quadratic$)$ functions as $x\to\pm\infty$. It is clear from their plots that for every even-order polynomial there is a big-enough free term $($positive or negative$)$ such that the graphic will no longer intersect the horizontal axis, meaning that our poly-nomial will have no roots. But this does not apply to odd-order polynomials, which always have at least one real root, since they always span from $-\infty$ to $+\infty$, or vice-versa.
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