Question

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

Answer 1

The solution is `x in (-oo,2]`

Explanation:

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

`f(2)=8+8-8-8=0`

Therefore,

`(x-2)` is a factor of `f(x)`

To find the other factors, perform a long division

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

`color(white)(aaaa)` `x^3-2x^2` `color(white)(aaaaaaaaaaaa)` `|` `x^2+4x+4`

`color(white)(aaaaa)` `0+4x^2-4x`

`color(white)(aaaaaaa)` `+4x^2-8x`

`color(white)(aaaaaaaaaa)` `0+4x-8`

`color(white)(aaaaaaaaaaaa)` `+4x-8`

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

So,

`f(x)=(x-2)(x^2+4x+4)=(x-2)(x+2)^2`

We build a sign chart

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

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

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

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

Therefore,

`f(x)<=0` when `x in (-oo,2]`

graph{x^3+2x^2-4x-8 [-21.2, 19.34, -12.25, 8.03]}