Question

How do you solve `x(x-1)(x+2)>0`?

Answer 1

The answer is `x in ] -2,0 [ uu ] 1,+oo [`

Explanation:

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

We do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaa)` `0` `color(white)(aaaa)` `1` `color(white)(aaaa)` `+oo`

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

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

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

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

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

graph{x(x-1)(x+2) [-8.89, 8.89, -4.444, 4.445]}

Answer 2

The solution for the inequality is `-2 < x < 0` or `x > 1`.

Explanation:

The inequality given is `x(x-1)(x+2)>0` i.e. product of all the terms is positive. It is apparent that sign of terms `(x+2)`, `x` and `(x-1)` will change around the values `-2`, `0` and `1` respectively. In sign chart we divide the real number line using these values, i.e. below `-2`, between `-2` and `0`, between `0` and `1` and above `1` and see how the sign of `x(x-1)(x+2)` changes.

Sign Chart

`color(white)(XXXXXXXXXXX)-2color(white)(XXXXX)0color(white)(XXXXX)1`

`(x+2)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XX)+ive color(white)(XXX)+ive`

`xcolor(white)(XXXXXXX)-ive color(white)(XXXX)-ive color(white)(XX)+ive color(white)(XXX)+ive`

`(x-1)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)-ive color(white)(XXX)+ive`

`x(x-1)(x+2)`
`color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive`

It is observed that `x(x-1)(x+2)> 0` when either `-2 < x < 0` or `x > 1`, which is the solution for the inequality.