is f irreducible

**georgel** · May 25th 2009, 09:37 AM

Is $f(x)=x^3+15x^2+8x+6$ irreducible in $\mathbb{Z}[x]$ , $\mathbb{Z}[[x]]$ , $\mathbb{Z}_5[x]$ ?

I've tried doing the $\mathbb{Z}[x]$ , but we've only mentioned Eisenstein's criterion and I can't apply it here.

Thank you.

**NonCommAlg** · May 25th 2009, 10:33 AM

Originally Posted by georgel

Is $f(x)=x^3+15x^2+8x+6$ irreducible in $\mathbb{Z}[x]$ , $\mathbb{Z}[[x]]$ , $\mathbb{Z}_5[x]$ ?

I've tried doing the $\mathbb{Z}[x]$ , but we've only mentioned Eisenstein's criterion and I can't apply it here.

Thank you.

the degree of $f$ is 3. so it's irreducible over $\mathbb{Z}$ iff it has no integer root. now suppose $x^3+15x^2+8x+6=0.$ then $x(x^2+15x+8)=-6$ and thus if $x \in \mathbb{Z},$ then we must have

$x=\pm 1, \pm 2, \pm 3, \pm 6.$ check these cases! the polynomial is reducible in $\mathbb{Z}[[x]]$ because for example: $f(x)=(1-x)(6+14x+29x^2 + 30x^3 + 30x^4 + 30x^5 + \cdots ).$

finally since $f(1) \equiv 0 \mod 5,$ we have $x-1 \mid f(x)$ in $\mathbb{Z}_5[x]$ and thus $f$ is reducible over $\mathbb{Z}_5.$

**georgel** · May 25th 2009, 11:12 AM

Originally Posted by NonCommAlg

the degree of $f$ is odd. so it's irreducible over $\mathbb{Z}$ iff it has no integer root.

I didn't know this. Could you possibly link the theorem and its proof, if it's available somewhere online?
Maybe it's obvious but I don't see it.

And thank you so much, again!

**georgel** · May 25th 2009, 11:41 AM

Originally Posted by NonCommAlg

the polynomial is reducible in $\mathbb{Z}[[x]]$ because for example: $f(x)=(1-x)(6+14x+29x^2 + 30x^3 + 30x^4 + 30x^5 + \cdots ).$

And I need further help with this as well.

How did you get this above?

Thank you if you have the time to help, if not, I'm already grateful for all your help!

**NonCommAlg** · May 25th 2009, 11:50 AM

Originally Posted by georgel

I didn't know this. Could you possibly link the theorem and its proof, if it's available somewhere online?
Maybe it's obvious but I don't see it.

And thank you so much, again!

sorry, i meant 3 not "odd". i don't know why i wrote odd?? well, there are only two ways to factorize a polynomial of degree 3: all factors of degree 1 or one degree 1 and one degree 2.

so either way, there will be a factor of degree 1 and therefore a root.

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