Question

How do you solve `(x-1)(x+2)(2x-5)<0` using a sign chart?

Answer 1

The answer is `x in ] -oo,-2 [ uu ] 1, 5/2[`

Explanation:

Let `f(x)=(x-1)(x+2)(2x-5)`

Now, we can do our sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaa)` `-2` `color(white)(aaaa)` `1` `color(white)(aaaa)` `5/2` `color(white)(aaaaa)` `+oo`

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

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

`color(white)(aaaa)` `2x-5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

`f(x)<0` when `x in ] -oo,-2 [ uu ] 1, 5/2[`