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Home Q & A Board Science-Related Homework Help Mathematics
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lessie_anicia lessie_anicia
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12 years ago
Prove that every polynomial of odd degree has a root?

Thanks in Advance.
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#1
12 years ago
The Fundamental Theorem of Algebra says that if a polynomial with a degree n >=1, then it has n roots.  We learned this in class today.  That means that quadratic have only 2 roots because it has a x^2
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#2
12 years ago
lim P(x) x==>-infinity is -infinity so it takes negative values
limP(x) x==> +infinity is +infinity so it takespositive values
 As P(x) is continuous for all x there is at least an x so P(x)=0
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#3
12 years ago
every polynomial of degree n (highest exponent) has n roots some of which can be complex numbers.
I think you are asking how to show that an odd polynomial has a real root.

If all the coefficients are real then complex roots come in pairs of the form
a (+/-) b i where i = sqrt(-1). If we factor out all the complex root pairs then the remaining must be real roots. And since we started with an odd numbered polynomial there must be at least one left.
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