Question 731699
Another root is 3-i, because complex roots of polynomial functions occur as conjugate pairs.  The corresponding quadratic factor for the function based on these two roots would be {{{(x-(3-i))(x-(3+i))=x^2-6x+10}}} [two steps omitted here].


Now knowing that quadratic factor, divide {{{x^4-2x^3-x^2-38x+130}}} by {{{x^2-6x+10}}}, polynomial division.  That quotient you could then find zeros for using general solution to quadratic equation.


Doing polynomial division for {{{(x^4-2x^3-x^2-38x+130) / (x^2-6x+10)=x^2+4x+13}}}, which is not factorable, so....
through general solution to quadratic formula, roots of this are -2-3i and -2+3i.


Summary of the roots:
3+i
3-i
-2-3i
-2+3i