What should be substituted to X so that x(cube)+2x(squared)-6x-4=0?

Answer 1

x^3+2x^2-6x-4=0..
x^3+x^2+x^2-6x-4=0
x(x^2+x-6)+x^2-4=0
x(x+3)(x-2)+(x-2)(x+2)=0
(x-2)(x)(x+3)+(x-2)(x+2)=0
(x-2)(x^2+3x+x+2)=0
(x-2)(x^2+4x+2)=0. Hence, either x-2=0 or x^2+4x+2=0. i.e. x=2 or x^2+4x+2=0. If x is a positive no. then it can only be =2. If x can be negative, the answers would be got by solving x^2+4x+2=0 which gives x=-2-sq. root of 2 OR x= -2 plus sq. root of 2.

Answer 2

Can't you do your homework alone? But as the answer is 0, x must be negative.