#1**+1 **

First, let's expand out everything. I'm not going to show the steps, but we eventually get

\(2z^4-32z^3+204z^2-608z+706=-8\\2z^4-32z^3+204z^2-608z+714=0\)

Now, since every number is divisible by 2, we get factor out 2 to get

\({2\left(z^{4}-16z^{3}+102z^{2}-304z+357\right)}=0\)

Now, note that \(z^{4}-16z^{3}+102z^{2}-304z+357=(z^{2}-8z+17)(z^{2}-8z+21)\)

This means that we have the equation

\(2\left(z^{2}-8z+17\right)\left(z^{2}-8z+21\right)=0\)

Since the equation is equal to 0, we get two equations. We have

\(z^{2}-8z+17=0,\qquad z^{2}-8z+21=0\)

Using the quadratic equation on both formulas, we find 4 values of z.

We have

\(z=4+i\\ z=4-i\\ z=4+\sqrt{5}\cdot (i)\\ z=4+\sqrt{5}\cdot (-i)\)

This was a very basic explenation of the problem.

I can go in dephth more if you want to.

Thanks! :)

NotThatSmart Jul 22, 2024