Question

3{x}^{4}-15{x}^{2}+12=0

Answer 1

$3x^4-15x^2+12=0\qquad|:3\\\\x^4-5x^2+4=0\\\\x^4-x^2-4x^2+4=0\\\\x^2(x^2-1)-4(x^2-1)=0\\\\(x^2-1)(x^2-4)+0\iff x^2-1=0\ \vee\ x^2-4=0\\\\x^2=1\ \vee\ x^2=4\\\\x=\pm\sqrt1\ \vee\ x=\pm\sqrt4\\\\\boxed{x=-1\ \vee\ x=1\ \vee\ x=-2\ \vee\ x=2}$