Hello can you help me with this problem?
Let $ p(x) = x^5 + x^4 + x^3 + 2x + 2 \in \mathbb{Z}_5 [x].$ Construct the field $\mathbb{F} = \mathbb{Z}_5 [x] \setminus \langle x^3 + 2x + 1 \rangle $ and write $ p(x) $ as a product of irreducible (over $ \mathbb{F}$) factors.
We know that $\mathbb{F} = \mathbb{Z}_5 [x] \setminus \langle x^3 + 2x + 1 \rangle $ So if we suppose that the root of $x^3 + 2x + 1 $ is $\alpha$ we will get: $$ \alpha^3 + 2 \cdotp \alpha + 1 = 0 $$ $$ \Downarrow $$ $$ \alpha^3 = -\left( 2 \cdotp \alpha + 1 \right) $$ Now let's find $ \alpha^4 $ and $\alpha^5$
$\alpha^3 = -\left( 2 \cdotp \alpha +1\right) $
$\alpha^4 = \alpha^3 \cdotp \alpha = -\left( 2 \cdotp \alpha +1\right) \cdotp \alpha = -\left( 2 \cdotp \alpha^2 +\alpha\right) $
$\alpha^5 = \alpha^4 \cdotp \alpha = -\left( 2 \cdotp \alpha^2 +\alpha\right) \cdotp \alpha = -\left( 2 \cdotp \alpha^3 +\alpha^2\right) = 4 \cdotp \alpha - \alpha^2 + 2$
$$ \Downarrow $$ $$ p(\alpha) = 4 \alpha - \alpha^2 + 2 -2\alpha^2 - \alpha - 2 \alpha - 1 = -3\alpha^2 + \alpha + 1 $$
Is this solution right?