Given one root is x1 = -2 + i.

Since the polynomial $P(x) = x^4 + 4x^3 + 6x^2 + 4x + 5$ has real coefficients, if a complex number is a root, its complex conjugate must also be a root.
Given one root is $x_1 = -2 + i$.
Therefore, its conjugate $x_2 = -2 - i$ is also a root.

A quadratic factor corresponding to these roots is $(x - x_1)(x - x_2)$.

This is in the form $(a-b)(a+b) = a^2 - b^2$, where $a = (x+2)$ and $b = i$.

We perform polynomial long division of $P(x) = x^4 + 4x^3 + 6x^2 + 4x + 5$ by $(x^2 + 4x + 5)$.

The quotient is $x^2 + 1$.
So, $P(x) = (x^2 + 4x + 5)(x^2 + 1)$.

Set the remaining factor equal to zero:

So, the other two roots are $x_3 = i$ and $x_4 = -i$.

The roots of the polynomial equation are $-2+i$, $-2-i$, $i$, and $-i$.

The roots of the polynomial equation are $-2+i, -2-i, i, -i$.