Question

How to solve `x^3-3x-2=0` ?

Answer 1

The roots are `-1,-1,2`

Explanation:

It is easy to see by inspection that `x = -1` satisfies the equation :

`(-1)^3-3times(-1)-2 = -1+3-2=0`

To find the other roots let us rewrite `x^3-3x-2` keeping in mind that `x+1` is a factor:

`x^3-3x-2 = x^3+x^2-x^2-x-2x-2`
`qquadqquad = x^2(x+1)-x(x+1)-2(x+1)`
`qquadqquad = (x+1)(x^2-x-2)`
`qquadqquad = (x+1)(x^2+x-2x-2)`
`qquadqquad = (x+1){x(x+1)-2(x+1)}`
`qquadqquad = (x+1)^2(x-2)`

Thus, our equation becomes

`(x+1)^2(x-2)=0`

which obviously has roots `-1,-1,2`

We can also see it in the graph:

graph{x^3-3x-2}

Answer 2

`x_1=x_2=-1` and `x_3=2`

Explanation:

`x^3-3x-2=0`

`x^3+1-(3x+3)=0`

`(x+1)(x^2-x+1)-3(x+1)=0`

`(x+1)(x^2-x+1-3)=0`

`(x+1)(x^2-x-2)=0`

`(x+1)(x+1)(x-2)=0`

`(x+1)^2*(x-2)=0`

Thus `x_1=x_2=-1` and `x_3=2`