Find an irreduciable polynomial in Q[x] (rational) which has exactly 5 real &2 complex roots?

Answer 1

x²+1 = 0 has 2 complex roots.
x²-1 = 0 has 2 real roots
x²-2 = 0 has 2 real roots
x-3 = 0 has 1 real root, so

(x²+1)(x²-1)(x²-2)(x-3) = 0 has the 5 real and 2 complex roots.

(x^4 - 1)(x^3 - 3x² - 2x + 6) = 0
x^7 - 3x^6 - 2x^5 + 6x^4 - x^3 + 3x² + 2x - 6 = 0

Answer 2

I disagree with the last answer. her polynomial is reducible over the rationals. In fact, she has given its reduction to lower-order factors. Got to do some work now, but I'll come back.