Answer
$x\in \{-2, -i, i, 1-\sqrt {3}i, 1+\sqrt {3}i\}$
Work Step by Step
$P(x)=x^5+x^3+8x^2+8$, factoring the polynomial,
$x^3(x^2+1)+8(x^2+1)=(x^3+8)(x^2+1)=(x+2)(x^2-2x+4)(x^2+1)$,
$x+2=0,x=-2$ or $x^2+1=0, x=\pm i$
solving for the trinomial using quadratic formula for the quadratic function of $ax^2+bx+c, x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
In this case $x^2-2x+4$, $\frac{2\pm \sqrt {(-2)^2-4\times1\times4}}{2\times1}=\frac{2\pm 2\sqrt {3}i}{2}=1\pm\sqrt {3}i$.
thus,
$x\in \{-2, -i, i, 1-\sqrt {3}i, 1+\sqrt {3}i\}$
