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I am trying to prove the irreducibility of $x^3 + 9x + 6 $ in $\mathbb{Q}[x]$ without using Eisenstein's criterion. What I have done is --

Let assume it is reducible in $\mathbb{Q}[x]$, then it can be written as $$x^3 + 9x + 6 = (a'x^2 + b'x+ c')(d'x + e')$$ $$= a'd'(x^2 + bx + c)(x + e)$$ $$\displaystyle= (x^2 + bx +c)(x+e)$$ $$\displaystyle= x^3 + (b+ e)x^2 + (be + c)x+ ce$$ equating the coefficients of same powers in both sides $$ b = -e, be+c = 9, ce = 6$$ solving these gives a equation in e $$e^3 + 9e - 6 = 0$$

I am stuck here, I don't know how to proceed from here.

Andrew Miller
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    If you can't use Eisenstein, can you use the rational roots theorem? It's probably easiest since the polynomial is a cubic. – Ben West Aug 20 '14 at 20:06
  • Well, I want to find out the problem with above argument. – Andrew Miller Aug 20 '14 at 20:10
  • The problem with the above method is exactly what you see - you end up with the same equation you started with. (I think you got it wrong, I think it ought to be $e^3+9e-6=0$, Why? Because $-e$ is a root of $x^3+9x+6$... – Thomas Andrews Aug 20 '14 at 20:17

3 Answers3

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I think you are well on your way to proving that any rational root of the equation must be an integer - and you ought to be able to identify the few possibilities for integer roots and test them. Since there must be a linear factor, this concludes the proof.

Mark Bennet
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Suppose $\frac{a}{b}$ is a root, where $a,b$ are relatively prime integers and without loss $b\ge 1$. Then $$(a/b)^3+9(a/b)+6=0$$ and hence $$a^3+9ab^2+6b^3=0$$ which we rearrange as $$a^3=b^2(-9a-6b)$$ Hence $b^2|a^3$, which is a contradiction unless $b=1$. Hence all rational roots are integers. Now the polynomial is strictly positive for $x\ge 0$, and strictly negative for $x\le -1$. Hence there are no integer roots.

vadim123
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    It is worth mentioning that this is basically the proof of the rational root theorem applied to this case – Belgi Aug 20 '14 at 20:28
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First: Your equation for $e$ is wrong. $$6=ce = e(9-be)=e(9+e^2)$$ so $e^3+9e-6=0$.

Second: This technique will always get back (essentially) the same polynomial you started with, so it is generally not a helpful helpful technique.

Thomas Andrews
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