## Algebra 2 (1st Edition)

$x=\pm\sqrt3$ or $x=2$ or $x=-2$
$x^6-4x^4-9x^2+36=(x-2)(x+2)(x^2-3)(x^2+3)=0$ Thus $x^2+3=0$ (but this is impossible since $x^2$ is non-negative) or $x^2-3=0\\x^2=3\\x=\pm\sqrt3$ or $x-2=0\\x=2$ or $x+2=0\\x=-2$