Question

How do you solve `x^3+2x^2-x-2>=0`?

Answer 1

The answer is `x in [-2,-1] uu [1,+ oo[`

Explanation:

Let `f(x)=x^3+2x^2-x-2`

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

So, `(x-1)` is a factor of `f(x)`

To find the other factors, we do a long division

`color(white)(aaaa)` `x^3+2x^2-x-2` `color(white)(aaaa)` `∣` `x-1`

`color(white)(aaaa)` `x^3-x^2` `color(white)(aaaaaaaaaaaa)` `∣` `x^2+3x+2`

`color(white)(aaaa)` `0+3x^2-x`

`color(white)(aaaaaa)` `+3x^2-3x`

`color(white)(aaaaaaaa)` `+0+2x-2`

`color(white)(aaaaaaaaaaaa)` `+2x-2`

`color(white)(aaaaaaaaaaaaa)` `+0-0`

Therefore,

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

So,

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

Now, we can construct the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `1` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)>=0`, when `x in [-2,-1] uu [1,+ oo[`