Question

What are relatively prime multinomials and how do they relate to factorability?

Answer 1

It depends. See explanation...

Explanation:

A polynomial with integer coefficients that cannot be factored into lower order polynomials with integer coefficients is a prime polynomial. This is not quite the same as not 'factorable' though, since you may still have a separable scalar factor:

e.g. `3x+6 = 3(x+2)`

A polynomial with coefficients in a set `S` is completely factored over `S` if it cannot be further divided into factors with coefficients in `S`. So if `S = QQ`, then you would be talking about rational coefficients and you would say that a polynomial is completely factored (over `QQ`) if it cannot be split into simpler factors with rational coefficients.

For example `2x^2-3y^2` is completely factored over `QQ` but not over `RR`, since it can be factored as:

`2x^2-3y^2 = (sqrt(2)x-sqrt(3)y)(sqrt(2)x+sqrt(3)y)`

This consideration of which set of numbers you can use affects both whether you consider a polynomial to be 'factorable' and whether you consider it to be 'completely factored'.

Any polynomial in one variable will have its full complement of roots in the complex numbers `CC`. So if you have a polynomial in one variable of order `n`, then it will have `n` (not necessarily distinct) roots in `CC` and a corresponding factoring into `n` linear factors with complex coefficients.

Given a polynomial `f(z) = a_nz^n + a_(n-1)z^(n-1)+...+a_1z+a_0`
there are `n` (possibly repeated) roots of `f(z) = 0`
`r_1, r_2, ... r_n in CC`, such that

`f(z) = a_n(z-r_1)(z-r_2)...(z-r_n)`

So you could say that a polynomial in one variable is not completely factored until it is reduced to a scalar multiple of linear factors.

In our context 'multinomial' means the same as 'polynomial' and is unrelated to whether the polynomial is 'factorable'.

'Relatively prime' talks about two polynomials having no common non-trivial factor. It is not applicable to whether a single polynomial is 'factorable'.