What is the easy way to calculate the roots of $z^4+4z^3+6z^2+4z$? What is the easy way to calculate the roots of $z^4+4z^3+6z^2+4z$? 
I know its answer: 0, -2, -1+i, -1-i. 
But I dont know how to find? Please show me this. I know this is so trivial, but important for me. Thank you. 
 A: $(z+1)^4=z^4+4z^3+6z^2+4z+1$.${}{}{}{}{}$
A: First, find the real rational roots.  One of them is $0$ because $z$ divides the polynomial.  Divide by $z-0 = z$. For the rest, look at the remaining polynomial $z^3 + 4z^2 + 6z + 4$. Its rational roots have numerator dividing $4$, the integer coefficient, and denominator dividing $1$, the coefficient of the highest term.  The sign is not restricted. So, the possibilities are $\pm 1, \pm 2,\ \text{and}\ \pm 4$. Trial and error gives $-2$ only.  Divide now by $z - (-2) = z+2$.  Now you have a quadratic.
A: Add 1 to both side to get:
$$z^4+4z^3+6z^2+4z+1 = 1$$
i.e.
$$(z+1)^4 = 1$$
can you finish from there?
A: Hint: Add $1$, and use the binomial theorem, for
$$z^4 + 4z^3 + 6z^2 + 4z + 1 = (z + 1)^4$$
So the problem is equivalent to solving $$( z+1)^4 = 1$$
A: put z4+4z3+6z2+4z = 0
z(z3 + 4z2+6z + 4) = 0
z = 0  (first root)

By hit n trial, -2 is also a root,
So divide z3 + 4z2+6z + 4 by (z+2) :
You will get another equation which can be solved using quadratic formula.
