Question

How do you solve `x^3+x^2-16x-16>0`?

Answer 1

`-4 < x <-1` or `x>4`

Explanation:

We have
`x^3+x^2-16x-16=(x-4)(x+1)(x+4)`
For
`x>4`
is this product positive.

The product also positive for

`-4 < x < -1`

Answer 2

The solution is `x in (-4,-1) uu(4,+oo)`

Explanation:

The inequality is

`x^3+x^2-16x-16>0`

Factorising,

`(x^2)(x+1-16(x+1))>0`

`(x^2-16)(x+1)>0`

`(x+4)(x-4)(x+1)>0`

Let `f(x)=(x+4)(x-4)(x+1)`

Build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `4` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x+4` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

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

`color(white)(aaaa)` `x-4` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

`f(x)>0` when `x in (-4,-1) uu(4,+oo)`

graph{x^3+x^2-16x-16 [-19.55, 26.05, -4.47, 18.34]}