Multiple roots of polynomials

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  • #1
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Homework Statement



We have this theorem:
Let [tex]f(x)\in F[x][/tex] Then f(x) has multiple roots if and only if
[tex]gcd(f(x),f'(x))=d(x)[/tex] and [tex]d(x)\geq 1[/tex]

We went BRIEFLY over the proof and we are supposed to be able to apply it on an upcoming exam.

I'm not exactly sure how it works or what I'm looking for.

Homework Equations





The Attempt at a Solution


To try to get a feel for it and see if it makes sense, I did this, because I know that it has multiple roots:
let [tex]f(x)=(x-1)^2=x^2-2x+1[/tex]
then[tex]f'(x)=2x-2[/tex]
so gcd(2,1)=1, therefore multiple roots?

Am I just looking at the degree of each one? Do I divide them? How does this theorem work ?

Any examples or clarification will be appreciated.
CC
 

Answers and Replies

  • #2
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Ok so I factor them just like integers, right?
so f'(x)=2(x-1)
and x-1 is the gcd.
Yes? No?
CC
 
  • #3
HallsofIvy
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Homework Statement



We have this theorem:
Let [tex]f(x)\in F[x][/tex] Then f(x) has multiple roots if and only if
[tex]gcd(f(x),f'(x))=d(x)[/tex] and [tex]d(x)\geq 1[/tex]
What do you mean by "gcd"? I would interpret it as "greatest common divisor" and that makes sense here because in this case f(x) and f'(x) must have a common factor. But then you set it equal to d(x) and say [itex]d(x)\geq 1[/itex] which makes no sense if d(x) is a polynomial. Do you mean that the degree of d(x) is greater than or equal to 0?

We went BRIEFLY over the proof and we are supposed to be able to apply it on an upcoming exam.

I'm not exactly sure how it works or what I'm looking for.

Homework Equations





The Attempt at a Solution


To try to get a feel for it and see if it makes sense, I did this, because I know that it has multiple roots:
let [tex]f(x)=(x-1)^2=x^2-2x+1[/tex]
then[tex]f'(x)=2x-2[/tex]
so gcd(2,1)=1, therefore multiple roots?
Am I just looking at the degree of each one? Do I divide them? How does this theorem work ?
No, of course, not. The derivative of any quadratic equation is of degree 1! That has nothing to do with multiple roots. I feel sure you mean "d(x)" to be a common factor. Here f(x)= (x-1)^2 and f'(x)= 2(x-1) so they have a common factor of x-1 which has degree equal to 1.
If you look at f(x)= x^2- 5x+ 6= (x-3)(x-2), f'(x)= 2x- 5 which has no common factor with f.

Any examples or clarification will be appreciated.
CC
If P(x) has a as a multiple-root, then P(x)= (x-a)nQ(x) where n> 1 and Q(x) is a polynomial. Now what is the derivative of P(x) using the product rule and the chain rule? Do you see that, for n> 0, P(x) and P'(x) must both have at least one factor of x-a? And so the degree of that common factor is greater than or equal to 1?
 
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  • #4
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yes yes yes
MUCH clearer. We did this in about 5 minutes. d(x), though we never really defined it has to be a common factor.
Another query:
If it doesn't factor, does that automatically mean it has no multiple roots?
for example
x^2+3x-1?

derivative is 2x+3, but no common factor.

If I have f(x)=x^50-1
then f'(x)=50x^49
So no common factors?

I hope the prof doesn't try to trick us, but he usually does.

Thanks
CC
 
  • #5
Hurkyl
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If it doesn't factor, does that automatically mean it has no multiple roots?
Yes and no. f(x) = x^4 + 2x^2 + 1, as a polynomial over R obviously doesn't have any real roots, so it certainly can't have a multiple real root. But f(x) and f'(x) have a nontrivial (real!) common factor, which represents the fact that f(x) does have multiple roots in C.

(Of course, this is all a moot point if F is an algebraically closed field)

You might want to ask your professor about that one; the question "Does f(x) have a multiple root in its splitting field over F" is a rather important one, and I can easily imagine that someone aksing "Does f(x) have a multiple root?" means that, and does not mean "Does f(x) have a multiple root in F?"
 
  • #6
HallsofIvy
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Any polynomial "factors"- into linear factors over the complex numbers or into linear or quadratic factors over the real numbers. In "elementary" algebra, factoring normally means factoring into terms having integer coefficients but that is not relevant here.
 

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