Is the polynomial $x^3 + 2x^2 + 1$ irreducible in $\mathbb{Z}_{17}[x]$?

Is the polynomial $x^3 + 2x^2 + 1$ irreducible in $\mathbb{Z}_{17}[x]$?

It seems this polynomial is reducible. How can I factor this? Thanks!

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You can just check all the values from $0$ to $16$ to see if it has a root. –  user66345 Apr 11 '13 at 4:25
Since it's cubic, if it factored it would have a linear factor, which means it'd have a root. Did you try testing for a root? If you find one, you can use long/synthetic division for the quotient, and then check if the resulting quadratic is irreducible or not by checking if the discriminant is a quadratic residue (and if it is, you can apply the quadratic formula). Otherwise, if this cubic has no roots, it's irreducible. (Caution - this does not apply to polynomials of degree greater than three, though.) –  anon Apr 11 '13 at 4:25
@user66345 so you're saying I can plug in the members of Z_17 to the function and see if it evaluates to 0? –  Allen Miller Apr 11 '13 at 4:28
@anon I tried 2 and found that it works. So that means I can use (x-2) as a root, right? Now I can use long division of (x-2)/ $x^3 + 2x^2 +1$ in Z_17? –  Allen Miller Apr 11 '13 at 4:34
One doesn't say $(x-2)$ is a root; one says $2$ is a root and $x-2$ is a factor. Yes, you do long division to find $(x^3+2x^2+1)/(x-2)$ in ${\bf Z}/17{\bf Z}$, and the long division thingie is written down starting with something that looks like $x-2\,|\overline{\,x^3+2x^2+1}$. By the way if you finish solving this problem, you can post your own solution to it. (In factor for homework or homework-type problems at this level, this is arguably the ideal situation.) –  anon Apr 11 '13 at 4:37
You've found that $2$ is a root of the cubic, and so it is not irreducible. If you want to factor it into irreducibles, then $$x^3+2x^2+1=(x-2)(x^2+ax+b)=x^3+(a-2)x^2+(b-2a)x-2b,$$ whence $a=4$ and $b=8.$ The only remaining question is whether $x^2+4x+8$ factors over $\Bbb Z_{17}.$ Noting that $$x^2+4x+8=(x+2)^2+4=(x+2)^2-13,$$ we find that it factors if and only if $13$ is a square modulo $17$. If it is, we can factor it as a difference of squares. (I leave it to you to determine.)
If $a$ is a root of your polynomial $f = x^3 + 2x + 1$, then $x - a$ divides $f$. Testing the elements of $\mathbb{Z}_{17}$, we find the roots $2$, $6$ and $7$. Hence $$x^3 + 2x + 1 = (x - 2)(x - 6)(x - 7).$$