## Algebra and Trigonometry 10th Edition

$2, -2, -1+\sqrt3i, -1-\sqrt3i$
$x^4+2x^3-8x-16 = x^3(x+2)-8(x+2)=(x^3-8)(x-2)=0$ Either x-2=0 giving x=2 or $x^3+8=0$ giving $(x+2)(x^2+2x+4)=0$, that gives either x+2=0 giving x=-2 or $x^2+2x+4=0$ $(x+1)^2+3=0$ $(x+1)^2 = -3$ x+1 = $+\sqrt3i$ or $-\sqrt3i$ Hence, x = 1$+\sqrt3i$ or $1-\sqrt3i$ for this case. Hence all in all x can be {$2, -2, -1+\sqrt3i, -1-\sqrt3i$}