# Multiplicity of roots

• Mar 25th 2007, 05:16 AM
Multiplicity of roots
I am trying to prove the following and don't even know where to get started:
f(x) = as sub n x^n+ a sub n-1 + ....+a0
f'(x) is n a sub n x ^ n-1 + (n-1) a sub n-1 x^n-2+...+a1. If u belongs to N and (x-c)^u divides f(x) but (x-c)^u-1 does not we can say that c is a root of f (x) multiplicity u
Now suppose c is a root of f (x) of multiplicity u greater than 1. Prove that c is a root of f'(x) as well and suppose c is root of both f (x) and conversely c is a root of f (x) of multiplicity u greater than 1.
• Mar 25th 2007, 05:33 AM
ThePerfectHacker
Quote:

I am trying to prove the following and don't even know where to get started:
f(x) = as sub n x^n+ a sub n-1 + ....+a0
f'(x) is n a sub n x ^ n-1 + (n-1) a sub n-1 x^n-2+...+a1. If u belongs to N and (x-c)^u divides f(x) but (x-c)^u-1 does not we can say that c is a root of f (x) multiplicity u
Now suppose c is a root of f (x) of multiplicity u greater than 1. Prove that c is a root of f'(x) as well and suppose c is root of both f (x) and conversely c is a root of f (x) of multiplicity u greater than 1.

It seems to be true.
I do not know if you are using a derivative or a polynomial operator which is like the derivative.

Meanining, when you write f'(x) do you mean the derivative or do you mean another polynomial (defined) as the derivative.

Let us assume you have f'(x) are the derivative.
Then you can write,
f(x)=R(x)*(x-c)^n
Because c is multiplicity of n.

But then,
f'(x)=R'(x)*(x-c)^n + R(x)*(x-c)^{n-1}
We see that (x-c)^{n-1} is a factor.
But (x-c)^{n-1} does not divide f'(x) because (x-c) does not divide R(x).
Thus, you assumption is true.
• Mar 26th 2007, 10:15 AM