Question

How do you solve and write the following in interval notation: `(x-4) (x+3) (2-x)<=0`?

Answer 1

The solution is `x in [-3,2] uu [4,+oo)`

Explanation:

Let

`f(x)=(x-4)(x+3)(2-x)`

We build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `2` `color(white)(aaaaa)` `4` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

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

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

Therefore,

`f(x)<=0` when `x in [-3,2] uu [4,+oo)`