A graph of the quartic reveals that there are two real and two complex roots.
One real root is close to -1, the other close to -4. Newton's iterative method:
zn+1=xn-f(zn)/f'(zn) gives us successive iterations for z which converge towards a solution. Two separate calculations in which z0=-1 and z0=-4 give the roots z=-0.833487 and -3.918241 approximately. f(z) is the given quartic, and f'(z) its first derivative.
If we call these real roots A and B then we can combine them into a quadratic:
(z-A)(z-B)=z2-(A+B)z+AB then divide it into f(z).
A+B=-4.751728, AB=3.265803.
z2+(4+A+B)z+(4+A+B)(A+B)-AB (quotient)
z2-(A+B)z+AB ) z4 +4z3 -z+1
z4-(A+B)z3 +ABz2
(4+A+B)z3 -ABz2 -z
(4+A+B)z3-(4+A+B)(A+B)z2 +AB(4+A+B)z
((4+A+B)(A+B)-AB)z2-(AB(4+A+B)+1)z
We need go no further because we have the required quadratic in the quotient: z2+(4+A+B)z+(4+A+B)(A+B)-AB.
This can be solved using the quadratic formula and substituting in the values of A and B, remembering that the roots are complex so we are expecting the square root of a negative quantity.
The quadratic is z2+px+q where p=4+A+B and q=(A+B)p-AB.
The roots are ½(-p±√(p2-4q)).
p=-0.751728, q=0.306203; √(p2-4q)=√-0.659719=0.812231i. So the two complex roots are 0.375864±0.406115.
So the four roots are: -0.833487, -3.918241, 0.375864+0.406115i, 0.375864-0.406115i (to 6 decimal places).