Show that for any real number a not 0, the polynomial x^n-a has no muktiple roots in the real numbers
Suppose p(x)=x^n-a
I'm not even sure what direction to go for this problem
If is a root of , prove that can't be a root of its derivative .
Fernando Revilla
You only need the formal derivative. That is an algebraic tool.
Fernando Revilla
f(x)=x^n-a
f'(x)=nx^(n-1)
I can do that, but am unsure about finding a gcd. I'm thinking we can use division to find the gcd.
(x^n-a)/(nx^n-1)=1/nx-a/nx^(n-1). So the remainder is -a.
But still not sure about the gcd since we want the gcd to not equal 1.