Is a^2 - 39 prime a) over the set of polynomials with rational coefficients B) real coefficients?

Answer 1

There is basically only one way to factor your polynomial:
a^2-39=(a-Sqrt(39))(a+Sqrt(39)).
(There are other ways to do it, but none of them are different enough to matter.)

Since Sqrt(39) is irrational, your polynomial can be factored over the reals but not over the rationals. So... (a) yes, (b) no.

Answer 2

Well, you have to ask yourself can a^2 - 39 be factored? Yes, so we have

(a + (39)^1/2) and (a - (39)^1/2)

Now, sqrt(39) is certainly a real number, so the polynomial factors in respect of real coefficients. But how about the rationals? Well, are there any rationals that can lead to their sum or difference being equal to sqrt(39)? No, since sqrt(39) = sqrt(3) x sqrt(13) and sqrt(3) is irrational. (So is sqrt(13) for that matter).

HTH

Charles