Seems easy, but how would you show that every polynomial of fifth degree has at least one real root?

Thanks for any help.

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- Nov 13th 2006, 05:09 AMOntarioStudPolynomial of fifth degree
Seems easy, but how would you show that every polynomial of fifth degree has at least one real root?

Thanks for any help. - Nov 13th 2006, 05:38 AMSoroban
Hello, OntarioStud!

Quote:

How would you show that every polynomial of fifth degree has at least one real root?

Graphically, maybe?

Given: .$\displaystyle f(x) \:=\:ax^5 + bx^4 + cx^3 + dx^2 + ex + f$

For positive $\displaystyle a: \;\lim_{x\to\infty}f(x) \,=\,\infty$ and $\displaystyle \lim_{x\to-\infty}f(x) \,=\,-\infty$*****

Since the polynomial function is continuous, it__must__cross the x-axis somewhere.

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*****.Of course, the argument hold for negative $\displaystyle a$ as well.

- Nov 13th 2006, 07:25 AMThePerfectHacker
In complex analysis it is shown that any non-constant polynomial is reducible to linear ploynomials over the complex field. Furthermore, the real polynomials are reducible to linear factors such that they come in conjugate pairs of complex numbers. Since there is an odd number of factors it is not possible to a polynomial into non real complex numbers. Thus the factor $\displaystyle (x-r)$ appears where $\displaystyle r$ is real. This assures us that there is at least one real solution.