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Math Help - No multiple roots in R

  1. #1
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    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
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by kathrynmath View Post
    Suppose p(x)=x^n-a I'm not even sure what direction to go for this problem
    If x_0 is a root of p, prove that x_0 can't be a root of its derivative .


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  3. #3
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    But why would I do anything with derivatives in an algebra course?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by kathrynmath View Post
    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
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  5. #5
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    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?
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  6. #6
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    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
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  7. #7
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    Quote Originally Posted by kathrynmath View Post
    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.
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  8. #8
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    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.
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  9. #9
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    The all roots of f'(x) are 0.
    Is 0 a root of f(x)?
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  10. #10
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    0 is a root of f(x)
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  11. #11
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    f(0) = 0^n - a = -a which is not 0,
    so 0 is not a root of f(x).
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  12. #12
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    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.
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  13. #13
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    Quote Originally Posted by kathrynmath View Post
    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
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