Question

$x^4-2x^2-12x-8=0$

Answer 1

$x^4-2x^2-12x-8=0 \x^4-2x^3+2x^3+4x^2-4x^2-2x^2-4x-8x-8=0 \(x^4-2x^3-2x^2)+(2x^3-4x^2-4x)+(4x^2-8x-8)=0 \x^2(x^2-2x-2)+2x(x^2-2x-2)+4(x^2-2x-2)=0 \(x^2+2x+4)(x^2-2x-2)=0 \ \1. \x^2+2x+4=0 \D=2^2-4*1*4=4-16=-12 \No...solutions \2. \2x^2-2x-2=0 \D=2^2-4*1*(-2)=4+8=12 \ \x_1=1+sqrt{3} \x_2=1-sqrt{3}$