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).

Re: Function is irreducible in the rationals.

Quote:

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.

And your LaTeX is good! :)

-Dan

Re: Function is irreducible in the rationals.

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".

Re: Function is irreducible in the rationals.