Thread: No multiple roots in R

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

Originally Posted by kathrynmath

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

But why would I do anything with derivatives in an algebra course?

Originally Posted by kathrynmath

But why would I do anything with derivatives in an algebra course?

You only need the formal derivative. That is an algebraic tool.


Fernando Revilla

I have this definition:
Let f(x) be in F[x]. An element c in F is said to be a root of multiplicity m>1 of f(x) if (x-c)^m|f(x) but (x-c)^(m+1) does not divide f(x).
Is there a way to use this definition somehow?

I also have that a nonconstant polynomial f(x) over the field R of real numbers has no repeated factors iff gcd(f(x),f'(x))=1

Originally Posted by kathrynmath

I also have that a nonconstant polynomial f(x) over the field R of real numbers has no repeated factors iff gcd(f(x),f'(x))=1

This is essentially what FernandoRevilla asked you to use in the first post.

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.

The all roots of f'(x) are 0.
Is 0 a root of f(x)?

0 is a root of f(x)

f(0) = 0^n - a = -a which is not 0,
so 0 is not a root of f(x).

Since the only root of f'(x)=0 and 0 is not a root of f(x), the gcd(f(x),f'(x))=1. Thus, no multiple roots.

Originally Posted by kathrynmath

Since the only root of f'(x)=0 and 0 is not a root of f(x), the gcd(f(x),f'(x))=1. Thus, no multiple roots.

Yup, just that simple