Equation $z^4+2z^3+6z^2+8z+8=0$ has the root $z = -2i$ Solve equation [closed]

The answer is $\pm 2i \$ and -$1 + - i$

Well another problem I am struggling with, not even sure how to start on this on, so can't even give out information on what I am trying to do.

closed as off-topic by Shaun, Nosrati, Travis, Did, Theoretical EconomistSep 3 '18 at 0:02

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• J.doe, was one of roots given? – user376343 Sep 2 '18 at 14:24
• @j.doe Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE – gimusi Oct 23 '18 at 20:50

Recall that since the coefficients are reals we have that $(z^2+4)$ divides $z⁴+2z³+6z²+8z+8=0$.
$$z⁴+2z³+6z²+8z+8=(z^2+4)(z^2+2z+2)$$
The equation in question has real coefficients. If $a+bi$ is a zero, so is $a-bi$, hence not only $-2i$ is a solution, but also $2i$. To find the other zeroes, divide the polynomial by $(z-2i)(z+2i)$ and solve the resulting quadratic.
Alt. hint: $\, + 2i\,$ must also be a root, and let $\,a,b\,$ be the other two roots. By Vieta's relations $\,-2i +2i + a + b=-2 \implies a+b=-2\,$ and $\,(-2i)(2i) a b = 8 \implies ab=2\,$. It follows that $\,a,b\,$ are the roots of the quadratic $\,z^2+2z+2=0\,$