Question

How do you solve `x^3+2x^2-4x-8>=0`?

Answer 1

The answer is `x>=2` or `x in [2,+oo[`

Explanation:

Let `f(x)=x^3+2x^2-4x-8`

Then `f(2)=8+8-8-8=0`
Therefore, `(x-2)` is a factor of `f(x)`

`f(-2)=-8+8+8-8=0`
Therefore, `(x+2)` is a factor of `f(x)`

So, `(x+2)(x-2)=x^2-4` is a factor of `f(x)`

To find the last factor, let's do a long division

`color(white)(aaaa)` `x^3+2x^2-4x-8` `color(white)(aaaa)` `∣` `x^2-4`

`color(white)(aaaa)` `x^3` `color(white)(aaaaaaa)` `-4x` `color(white)(aaaaaaa)` `∣` `x+2`

`color(white)(aaaa)` `0+2x^2` `color(white)(aaaaa)` `0-8`

`color(white)(aaaaaa)` `+2x^2` `color(white)(aaaaaaa)` `-8`

`color(white)(aaaaaaaa)` `0` `color(white)(aaaaaaaaaa)` `0`

So, `f(x)=(x+2)^2(x-2)`

As `(x+2)^2>0`

Therefore, the sign of `f(x)` will depend on the sign of `(x-2)`

When `x<2`, `f(x)<0`

and when `x>=2`, `f(x)>=0`

graph{x^3+2x^2-4x-8 [-20.28, 20.27, -10.14, 10.14]}