Talk:Rational root theorem
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In what sense are you dividing the set p by the set q? --RoseParks
That was an error in expression on my part. I was going to come back later on and fix it, but apparently someone else has done so already. --Apeiron
I removed the line
(±1 cannot be a root because a0/an is not integral.)
It so happens that neither 1 nor -1 is a root, but this is not implied by the rational root theorem. There is also no theorem saying that ±1 can only be a root if a0/an is integral. For example, 3x3-x2-x-1 = 0 has a root x=1, even though -1/3 is not integral.
I also simplified the statement about the fundamental theorem of algebra. The versions "has a root" and "has n roots, if counted with multiplicities" are equivalent (i.e., each one easily implies the other -- from one root you can get to n roots with the above-mentioned Horner scheme), but I think the first version is easier to comprehend. Anyway, the second version can be found when following the link to the fundamental theorem of algebra.
Aleph4 09:41, 14 Apr 2004 (UTC)
I added the stipulation that a0 should be nonzero, since if it is 0 then 0/1 is a rational root, but the numerator 0 is not a divisor of the constant term 0 - 0/0 is undefined.
