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!
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!
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 $f = x^3 + 2x^2 + 1$, then $x - a$ divides $f$. Testing the elements of $\mathbb{Z}_{17}$, we find the roots $2$, $6$ and $7$. Hence $$x^3 + 2x^2 + 1 = (x - 2)(x - 6)(x - 7).$$