# Does a cubic polynomial split in linear factors over the rational numbers?

Let $\alpha + \beta + \gamma = a$ , $\alpha \beta + \alpha \gamma + \beta \gamma = b$, $\alpha \beta \gamma = c$, $\alpha^2 \beta + \gamma^2 \alpha + \beta ^2 \gamma = d$, where $\alpha, \beta, \gamma$ are complex numbers and $a,b,c,d$ are rational numbers. Can it happen, that for example $\alpha$ is a rational number while $\beta, \gamma$ are irrational, or is this impossible? I guess that from the last equation it would follow that $d$ is not rational if only $\alpha$ is rational, but am unable to prove it. The polynomial I am considering is $f(x) = x^3 - ax^2 + bx -c = (x-\alpha)(x-\beta)(x-\gamma)$ with the constraint that $\alpha^2 \beta + \gamma^2 \alpha + \beta ^2 \gamma = d$ is rational.

If I don't misread your equations, and given the title, you are asking if a cubic $x^3 + a x^2 + b x + c$ can have one rational and two irrational roots, if it has rational coefficients. Just consider:

$$(x - 1) (x^2 - 2) = x^3 - x^2 - 2 x + 2$$

with one rational and two irrational roots.

• This example was clear to me, but the roots do not satisfy the rationality constraint of the number $d$.
– user276611
Jan 2 '16 at 7:05

Assume that $a,b,c,\alpha \in Q.$ Then $S=\beta+\gamma=a-\alpha\in Q$ and $P=\beta\gamma=b-(\beta+\gamma)\alpha\in Q.$ So $\{\beta,\gamma\}$ is the set of solutions of $y^2-S y+P=0$ with $P,S \in Q.\quad$ First, if $\beta=u+i v$ where $u,v$ are real and $v\ne 0$,then $\gamma=u-i v.$ Then $d\in Q\to Im( d)=0.$ But $Im(d)=0$ is a quadratic equation in $\alpha$ which has no real solution unless its discriminant $-4v^4$ is $0,$ a contradiction.$\quad$ Second, if $\beta =u+\sqrt v$ with $u,v\in Q$ and $0<v$ and $\sqrt v\not \in Q,$ then $\gamma=u-\sqrt v\;$. If $d\in Q$ then the co-efficient, in $d$, of $\sqrt v$, must be $0.$ But this co-efficient is $(a-u)^2-v$, implying $\sqrt v\in Q,$ a contradiction. Therefore if $a,b,c, \alpha \in Q$ then $(d\in Q)\to (\beta,\gamma\in Q)$........(Remark. $Im(d)$ is the imaginary part of $d$.)

• Thanks, nice answer. I started similar, but the computations got ugly and didn't finish. Now I was able to follow with a computer algebra system which I should have used in the first place for the computations.
– user276611
Jan 2 '16 at 7:56
• You're welcome.I think the first part can be omitted . The second part seems to cover it all. I would just re-write "$\beta=u+w$ where $u\in Q, w\not \in Q$ and $w^2\in Q$" and calculate the co-efficient of $w$ in $d$. Jan 2 '16 at 8:23

Suppose you put $\beta=\sqrt 2$ and $\gamma =-\sqrt 2$ so that $\beta+\gamma=0$ and $\beta\gamma=-2$, then your equations become $$\alpha = a$$$$b=-2$$ $$-2\alpha = c$$ and $$\alpha^2\sqrt 2+2\alpha-2\sqrt 2=d$$

Now that illustrates the problem you have, because the last equation would only be rational if $\alpha^2=2$

However we can choose $\beta$ and $\gamma$ as conjugate irrationals with rational sum and product $s=\beta+\gamma$ and $p=\beta\gamma$ in which case the first three equations are rational. The fourth, problem, equation is not homogeneous and becomes $$(\alpha^2+p)\beta+\gamma^2\alpha=d$$Here $\beta=s-\gamma$ so that $$\alpha\gamma^2-(\alpha^2+p)\gamma+\alpha^2 s+ps-d=0$$ and this is a rational quadratic for $\gamma$ whence $$s=\frac {\alpha^2 +p}\alpha; p=\alpha^2s+ps-d$$ From the first $p=\alpha s-\alpha^2$ so you get a quadratic for $s$ by substituting in the second which needs to have rational roots.

This is rather rambling and incomplete - supper interrupts, but it ought to be possible to complete from here.