I was trying this problem from my Abstract Algebra book exercise that says:

Show that the polynomial $x^2+\frac 13x-\frac 25$ is irreducible in $\mathbf Q[x]$.

What I tried: $x^2+\frac 13x-\frac 25 \equiv 15x^2+5x-6=f(x)$,say.
Now I compute ,$f(x+1)=15x^2+35x+14$. Now, $ 7 \mid 35, 7 \mid 14; 7\nmid 15,7^2 \nmid 14$. Hence using Eisenstein's criterion ,$f(x+1)$ is irreducible over $\mathbf Q$. Hence $f(x)$ is irreducible over $\mathbf Q$.

Am I right?

I want to add another problem similar to the above :

Show that the polynomial $2x^3-x^2+4x-2$ is not irreducible in $\mathbf Z[x]$.

Here,$f(x+1)=2x^3+5x^2+8x+3$ where $f(x)=2x^3-x^2+4x-2$. Here ,I can not apply Eisenstein's criterion for $f(x+1)$. I can not also factorize $f(x)$.Then how Can I prove it?

Can someone point me in the right direction? Thanks in advance for your time.


Your answer is:

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Post question's edit:

You want to prove that the polynomial is reducible in $\Bbb Z[x]$. By (the second) Gauss's lemma, if it is reducible in $\Bbb Q[x]$, then it also is in $\Bbb Z[x]$. Since its degree is $3$, it is reducible (over $\Bbb Q$) if, and only if, there is a rational root. Now note that in case Eistein's criterion doesn't work, you still won't know wether it is reducible or not, so that shouldn't be the way to go. You want to find a rational root. Try finding roots using the rational root theorem.

Something else you can try is to factor your polynomial. If you can't do this, you can always put the polynomial as input in a very well known on-line software.

  • $\begingroup$ thanks a lot @Git Gud.Got it. $f(\frac12)=0$ saves the day. $\endgroup$
    – learner
    May 22 '13 at 6:57

$\ \begin{eqnarray}{\rm Hint\!:}\quad\ &2x^3-x^2+4x-2 \\ = &\,x^2(2x-1)+ 2(2x-1)\end{eqnarray}$


It looks fine to me. You might also consider applying the rational root theorem to show that the polynomial has no roots in $\mathbb{Q}$, and therefore it is irreducible since it has degree less than three.


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