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kabutjv7 2022-05-01 Answered
Let f ( x ) be a monic polynomial of odd degree. Prove that A R s.t. f ( A ) < 0 and there exists B R such that f ( B ) > 0.

Deduce that every polynomial of odd degree has a real root.

There are questions that answer the final part, but they do not do so by proving the first part. I am fairly sure that this involves the intermediate value theorem, but not sure how to implement it in this case.
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Answers (1)

RormFrure6h1
Answered 2022-05-02 Author has 13 answers
For the final part. If you have f ( A ) < 0 and f ( B ) > 0 then by the IVT every value in [f(A),f(B)] is attained by f(x) for some x between A and B, and this includes 0.
To show the existence of the A and B show that for x large one has that the sign of f(x) is the sign of the leading coefficient. And, if the degree is odd for small x one has that the sign of f(x) is the opposite sign of the leading coefficient.
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