Answer
$-1,3,i,-i$
Work Step by Step
Consider $P(x)=x^4=2x^3-2x^2-2x-3=0$
In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows:
$x^4=2x^3-2x^2-2x-3=(x+1)(x^3-3x^2+x-3)=0$
This can be re-written as:$(x+1)[x(x^2+1)-3(x^2+1)]=0$
This gives: $(x+1)(x^2+1)(x-3)=0$
Here, $x+1=0 \implies x=-1\\x-3=0 \implies x=3\\x^2+1=0 \implies x^2=-1 \\ or, x=\pm i$
Thus the zeroes of $P(x)$ are: $-1,3,i,-i$