Question 1163519: Question: I always thought, till now, that the number of possible roots for a polynomial is equal(or atmost) to its highest degreee. Also I have been counting each radical and complex root as two since they all have +/- values (just like +/- n will be counted as two). Then I came across the following polynomials (degree four and five):
4X^4 +8X^3 -5X^2 -2X +1 = 0
3X^5 +2X^4 -15X^3 -10X^2 +12X +8 = 0
The second one as expected has 5 roots: +/-1, +/-2, -2/3
But the first one which (degree-4) has FIVE roots! viz., +/-1/2, +/-√2 and -1.
So which of assumption is wrong?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
What assumption? There are no assumptions in the problem as stated.
It is a fact that the number of roots of a polynomial is equal to the degree of the polynomial.
What is not true is your statement that the 4th degree polynomial has 5 roots -- it of course does not.
Specifically, -1 is not a root.
The roots are +/-.5 and -1+/-sqrt(2).
If you ever think you are finding a n-th degree polynomial with more than n roots, check the work you did to determine the roots, because something in your calculations has to be wrong.
Answer by ikleyn(52803)  (Show Source):
You can put this solution on YOUR website! .
It is a reminiscence of the Rational root theorem in your mind.
The Rational root theorem says that all the rational roots of a polynomial with integer coefficients
are among the fractions +/-  , where q is a divisor of the leading coefficient and p is a divisor of the constant term.
But this theorem DOES NOT state
a) NEITHER that all the roots of a polynomial are among these fractions
(irrational possible roots are DEFINITELY out of this set; complex roots ALSO are out of this set)
b) NOR that each root goes two times with the " + " and " - " sign.
About this theorem, read this Wikipedia article
https://en.wikipedia.org/wiki/Rational_root_theorem
and have fan (!)
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