Polynomial coefficient equivalence

what is the fundamental algebra result that allows you to equate coefficients for 2 polynomials if they have equal values for all values of x?

and how can you replace this expression-

F(x)=A(x-alpha1)(x-alpha2)....)x-alphar)

with F(x)=A(x-alpha1)^m1(x-alpha2)^m2....(x-alphar)^mr, with A=an

mk are integers >=1 and are multiplicities of the roots mk being the multiplicity of alphak

thanks very much!

Re: Polynomial coefficient equivalence

Re: Polynomial coefficient equivalence

If polynomial and polynomial are equal, for all x, then

for all x. In particular, taking x= 0, [itex]a_0= b_0[/tex]. Since those are equal, we can cancel them leaving

which is the same as

For x not 0, we can divide by x and have

and now, **because polynomials are continuous**, taking the limit as x goes to 0, .

Just repeat that to get for all k up to n.

Notice that I had to say "x not 0" at one step so that I could not just let x= 0 in the next. But because polynomials are continuous I can say instead "take the limit as x goes to 0".

Quote:

and how can you replace this expression-

F(x)=A(x-alpha1)(x-alpha2)....)x-alphar)

with F(x)=A(x-alpha1)^m1(x-alpha2)^m2....(x-alphar)^mr, with A=an

mk are integers >=1 and are multiplicities of the roots mk being the multiplicity of alphak

thanks very much!

You **can't**! Those are very different functions. Now, if you did NOT use "alpha1" to "alphar" in both, you **could** if some of the "alpha"s are the same, combine them. That is, if alpha1= alpha2, you could write (x- alpha1)(x- alpha2) as (x- alpha1)^2. But then you would not have the "(x- alpha2)" term.

Re: Polynomial coefficient equivalence

thanks so much for the replies! sorry I have been having serious health problems

this is from p. 4 of the Riley/Hobson math. methods:

the condition f(Ak)=0 for k=1,2,...r could be met if

F(x)=A(x-alpha1)(x-alpha2)....(x-alphar) were replaced by

F(x)=A(x-alpha1)^m1(x-alpha2)^m2....(x-alphar)^mr

Re: Polynomial coefficient equivalence

I know that for cubic equations you can solve by the partial differentiation method, but I'm sure you already know this...