(adsbygoogle = window.adsbygoogle || []).push({}); 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. [tex](x^2+a)(x+b)[/tex] 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 [itex]ab[/itex] 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 [itex]x[/itex] and [itex]x^2[/itex] 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 [itex]x^3+a=0[/itex] where a is rational and [itex]a^{1/3}[/itex] is not rational.

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# Homework Help: Roots of cubics

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