$x^2 +1$ is reducible over $\Bbb Z_5$ How is it true that the polynomial $x^2 +1$ is reducible over $\mathbb{Z}_{5}$?
Is it because  that 3 is an element of $\mathbb{Z}_{5}$ and $3^2 +1=10$ which is divisible by 5.
And so as Polynomial $x^2 +1$ having degree 2 has got a root in $\mathbb{Z}_{5}$.
$x^2 +1$ is reducible over $\mathbb{Z}_{5}$
Is this explanation reasonable?

You have found one of the roots, by inspection you will notice that $2$ is also a root. You have
$$x^2+1 = (x-2)(x-3)$$
if $\mathbb{Z}_5[x]$.
