Proof about an odd degree polynomial

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    Degree Polynomial Proof
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SUMMARY

A polynomial of odd degree in R[x] with no multiple roots must have an odd number of real roots. This conclusion is supported by the fact that complex roots occur in conjugate pairs, leading to an odd count of real roots remaining. The discussion clarifies that while every odd degree polynomial has at least one real root, it does not guarantee that all roots are real. The key takeaway is the relationship between the degree of the polynomial and the nature of its roots.

PREREQUISITES
  • Understanding of polynomial functions in R[x]
  • Knowledge of complex numbers and their properties
  • Familiarity with the Fundamental Theorem of Algebra
  • Basic concepts of root multiplicity
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  • Study the Fundamental Theorem of Algebra in detail
  • Explore the concept of root multiplicity in polynomials
  • Learn about complex conjugate pairs and their implications
  • Investigate examples of odd degree polynomials and their real roots
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chaotixmonjuish
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How could one sufficiently prove that a polynomial of odd degree in R[x] with no multiple roots must have an odd number of real roots?

My book just refers back to a Corollary that states every polynomial of odd degree in R[x] has a root in R. However it doesn't say, all roots are in R.
 
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chaotixmonjuish said:
However it doesn't say, all roots are in R.

That's right. You're not trying to prove that all the roots are in R, merely that an odd number of them are. (You actually can't prove that all the roots are real. It's not hard to come up with a cubic equation with exactly one real root.)

One thing to ask yourself is whether complex roots have to come in pairs.
 
Well I'm guessing its because all the complex roots will come as conjugate pairs, i.e. there are an even number of complex roots, so there will be an odd number of real roots left over.
 

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