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.

Solution 1:

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?

Solution 2:

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 $$( z+1)^4 = 1$$

Solution 3:

$(z+1)^4=z^4+4z^3+6z^2+4z+1$.${}{}{}{}{}$

Solution 4:

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.

Solution 5:

put z⁴+4z³+6z²+4z = 0
z(z³ + 4z²+6z + 4) = 0
z = 0 (first root)