Question 976056
Another root is x=i, because complex roots for polynomial functions come as conjugate pairs.  Those two roots are as  {{{(x-(-i))(x-i)=(x-i)(x+i)}}}
{{{x^2-i^2}}}
{{{x^2-(-1)}}}
{{{highlight_green(x^2+1)}}}, a factor of the polynomial p(x).


Use polynomial division for   {{{x^4+4x^3+6x^2+4x+5}}} as dividend
and {{{x^2+1}}}  as divisor.  The quotient represents the rest of the factors, as quadratic, degree two.
-
The division process not shown here; but result is {{{x^2+4x+5}}} as the quotient.


Use general solution method for a quadratic equation to find the zeros of this factor:
roots are
{{{x=(-4+- sqrt(4^2-4*5))/2}}}
{{{x=(-4+- sqrt(-4))/2}}}

{{{x=(-4+- 2i)/2}}}

{{{highlight(x=-2+- i)}}}   or   -2-i  and  -2+i.


(along with  -i  and i ).