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As we know, every polynomial with real coefficients of odd degree has at least one real root. We also know that real numbers are regarded as complex numbers $a+0i$ (where the imaginary part is zero).

Do odd degree polynomials have all complex roots?

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  • $\begingroup$ Complex roots come in conjugate pairs $\endgroup$ – akech Dec 15 '15 at 3:58
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    $\begingroup$ @DanielAkech Polynomial may have complex coefficients... OP has said nothing about that. Then complex roots need not come in conjugate pairs. $\endgroup$ – SchrodingersCat Dec 15 '15 at 4:00
  • $\begingroup$ You are correct, he will have to specify that. $\endgroup$ – akech Dec 15 '15 at 4:01
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Do odd degree polynomials have all complex roots?

Hint: If that were the case, then there would be no real root, meaning that the graphic of the function would never cross the horizontal axis. But a polynomial of odd degree is a continuous function which tends towards positive infinity at one end, and towards negative infinity at the other. But going from a positive value to a negative one, or vice-versa, without passing through zero, would contradict continuity.

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The Fundamental theorem of algebra states that :

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with an imaginary part equal to zero.

Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed.

The theorem is also stated as follows as you have remarked in your question:

Every non-zero, single-variable, polynomial of degree $n$ with complex coefficients has, counted with multiplicity, exactly $n$ roots. The equivalence of the two statements can be proven through the use of successive polynomial division.

So as keeping with what you are saying, we can say a polynomial of any degree $n$, irrespective of whether $n$ is odd or even and what nature the coefficients of the polynomial are, has $n$ complex roots.

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  • $\begingroup$ @Parag Gupta: Just to clarify Aniket's answer. Any $n$-th deg polynomial with real coefficients have $n$ complex roots $a+bi$. However, if $n$ is odd, then there are an odd number of roots with $b = 0$. $\endgroup$ – Tito Piezas III Dec 15 '15 at 4:31

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