if a polynomial equation of degree n has exactly one distinct

someone pls help

>>8737764
it's a non-negative integer

>>8737768
why,

explain the reasoning pls

>>8737773
its non-negative because it has a root

>>8737773

you just got it to feel it bro

In all seriousness, write out a couple of simple quadratic examples and see for yourself.

>>8737774
-1 has a root (i)

>>8737796
-1 is a constant polynomial, it doesn't have any roots

n is odd. If n was even, there would be at least two real roots as imaginary roots exist in complex pairs

>>8737810
He said real root. What we know is that either there is two roots or one root of multiplicity two i think.

>>8737818
multiplicity does not need to be 2

see: x^4 = 0

>>8737810
wrong

(x-1)(x-i) has one distinct real root

>>8737810
>(x-1)(x-i)
only if the coefficients are all real.

>>8737764
(x-a)^n will always have one distinct root (x=a) for positive values of n. Technically n doesn't even have to be an integer, though by definition polynomials have nonnegative integer exponents. So you can't conclude anything that wasn't already given by "a polynomial equation".

>>8737800
What about 0, it's constant polynomial too, but it has infinitely many roots

>>8737764
N is odd, polynomials have real coefficients and integer exponents.

>>8738795
it doesn't have a distinct real root so it's not relevant to the question

>>8739030
>N is odd, polynomials have real coefficients and integer exponents.
???

polynomials can have coefficients in any ring, you certainly can't say (x-i) isn't a polynomial

>>8739030
>polynomials have real coefficients
Underage b&