# Thread: Function is irreducible in the rationals.

1. ## Function is irreducible in the rationals.

Is $\displaystyle a=3x^3+2x+2$ reducible in the rationals?

So $\displaystyle a$ is reducible iff we can write $\displaystyle a=bc$, for $\displaystyle bc$ in the rationals (polynomials) and b,c$\displaystyle \neq$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)$\displaystyle \neq$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).

2. ## Re: Function is irreducible in the rationals.

Originally Posted by A6363
Is $\displaystyle a=3x^3+2x+2$ reducible in the rationals?

So $\displaystyle a$ is reducible iff we can write $\displaystyle a=bc$, for $\displaystyle bc$ in the rationals (polynomials) and b,c$\displaystyle \neq$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)$\displaystyle \neq$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).
You've got a handle on things. If there are no rational roots to a(x) = 0, then it cannot be factored over the rationals. In fact, that's all you need to do. You don't need to worry about the degrees.

The "rational root theorem" will be very useful here: if the polynomial equation, $\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$, has rational root $\displaystyle x= \frac{m}{n}$, where m and n are integers with no common factors, then the numerator, m, must evenly divide the constant term, $\displaystyle a_0$, and the denominator, n, must evenly divide the leading coefficient, $\displaystyle a_n$. 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".