# Homework Help: Roots of cubics

1. Jan 9, 2010

### Mentallic

1. The problem statement, all variables and given/known data
Not so much a homework problem, but a problem that is annoying me because of its simplicity.

Not all cubic polynomials with rational coefficients can be factorized by the rational root theorem (or is this false?). What I am finding hard to comprehend is how a cubic with only irrational or imaginary roots can have all rational coefficients.
What I'm looking for is an example or two illustrating a completely factorized cubic with non-rational roots that, if expanded, has all rational coefficients.

3. The attempt at a solution
In an attempt to figure out how it can be done myself I have realized that it is impossible to have one quadratic factor (rational coefficients) that does not have rational roots and the last factor still satisfying this.

e.g. $$(x^2+a)(x+b)$$ given a>0, the quadratic factor will have imaginary coefficients and of course b will need to be irrational or imaginary, but then if we expand this, the constant will be $ab$ which - if a is rational - will not satisfy the criteria, or if a is such a number that ab becomes rational then we will end up breaking the criteria again since the coefficients of $x$ and $x^2$ will be b and a respectively (which we've already shown have to both be non-rational).

I've tried a few others, but I'm sure you get the point

And of course this problem of mine applies to all polynomials of odd degrees.

EDIT: Oh and please exclude the obvious (for me anyway) cases of $x^3+a=0$ where a is rational and $a^{1/3}$ is not rational.

2. Jan 9, 2010

### Dick

I'm not sure why you are excluding x^3+a=0, but ok, this particular example came up in a problem earlier. 8x^3-6x-1=0 has three real roots. None of them are rational. By the rational roots test. The roots are cos(pi/9), and (1/2)*(-cos(pi/9)+/-sqrt(3)*sin(pi/9)). Kind of complicated, but what did you expect? I still like x^3-2=0 much better.

Last edited: Jan 10, 2010
3. Jan 10, 2010

### Mentallic

I excluded that obvious type because I know there were more cases where the coefficients of x and x2 weren't zero, and that one is really simple of course.
Thanks, I see where they are coming from now.