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

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- Jan 8th 2011, 07:28 AMkathrynmathNo 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 - Jan 8th 2011, 08:21 AMFernandoRevilla
If $\displaystyle x_0$ is a root of $\displaystyle p$, prove that $\displaystyle x_0$ can't be a root of its derivative .

Fernando Revilla - Jan 8th 2011, 08:44 AMkathrynmath
But why would I do anything with derivatives in an algebra course?

- Jan 8th 2011, 08:47 AMFernandoRevilla
You only need the

*formal*derivative. That is an algebraic tool.

Fernando Revilla - Jan 8th 2011, 09:46 AMkathrynmath
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? - Jan 8th 2011, 09:50 AMkathrynmath
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

- Jan 8th 2011, 10:24 AMsnowtea
- Jan 8th 2011, 11:20 AMkathrynmath
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. - Jan 8th 2011, 11:26 AMsnowtea
The all roots of f'(x) are 0.

Is 0 a root of f(x)? - Jan 8th 2011, 11:31 AMkathrynmath
0 is a root of f(x)

- Jan 8th 2011, 11:37 AMsnowtea
f(0) = 0^n - a = -a which is not 0,

so 0 is not a root of f(x). - Jan 8th 2011, 11:44 AMkathrynmath
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.

- Jan 8th 2011, 11:46 AMsnowtea