Is reducible in the rationals?
So is reducible iff we can write , for in the rationals (polynomials) and b,c 0. So deg(a)=3, therefore 0<deg(b)<deg(a) and 0<deg(c)<deg(a). So either deg(a)=2 or deg(a)=1, since deg(b)+deg(c)=deg(a)=3
The next step I'm not too sure about, but I believe it is to show that a(x) 0 for any x in the rationals. And this leads to the conclusion that a is irreducible since we cannot write a=bc since either b or c is a unit.
Any help will be appreciated, sorry that my notation is not entirely accurate in this post (not used to LateX).
The "rational root theorem" will be very useful here: if the polynomial equation, , has rational root , where m and n are integers with no common factors, then the numerator, m, must evenly divide the constant term, , and the denominator, n, must evenly divide the leading coefficient, . Here, that means that the numerator must be one of 1, -1, 2, or -2 while the denominator must be one of 1, -1, 3, or -3. That gives 1, -1, 2, -2, 2/3, and -2/3 as the only possible rational roots. Try them and if none actually are roots of the equation, it has no rational roots and the polynomial is "irreducible over the rationals".