Is $x^3+4$ irreducible in $Z_{11}[x]$?

Question

What are the factors if it is not? I am not sure how to figure this out in regards to $Z_{11}[x]$.

Answer 1

Over any field, a cubic is irreducible if and only if it has no roots.

You can just check whether this has any roots over $\mathbb{Z}_{11}$.

As TMM points out in the comments, $6$ is a root. Therefore $x^3+4$ is reducible over $\mathbb{Z}_{11}$.

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You can use the divison algorithm (remembering that you are working mod $11$) to get that:

$$ x^3 + 4 \equiv (x-6)(x^2+6x + 3). $$

You can then see if $x^2+6x + 3$ has any roots to factor further.