Question

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

Answer 1

The expression is `>=0` for `x>2`

Explanation:

let `f(x)=x^3+2x^2-4x-8`
Let us start by factorising the expression
By trial and error, ·`(x-2)` is a factor as `f(2)=0`

let us make the long division
`x^3+2x^2-4x-8` `color(white)(aaa)` `∣` `x-2`
`x^3-2x^2` `color(white)(aaaaaaaaaaa)` `∣` `x^2+4x+4`
`0+4x^2-4x`
`color(white)(aaaa)` `4x^2-8x`
`color(white)(aaaaa)` `0+4x-8`
`color(white)(aaaaaaaaa)` `4x-8`
`color(white)(aaaaaaaaaa)` `0+0`

so `f(x)=(x-2)(x^2+4x+4)=(x-2)(x+2)^2`
`f(x)>=0`
Looking at the factorisation
`(x+2)^2>=0` for all values of `x∈RR`
`(x-2)>=0` for `x>=2`
So the answer is `x>=2`