Question

How do you solve and write the following in interval notation: `(x^2-5x+6)/(x+3) >0`?

Answer 1

The solution is `x in (-3,2) uu (3,+oo)`

Explanation:

We cannot do crossing over

Factorise the inequality

`(x^2-5x+6)/(x+3)>0`

`((x-2)(x-3))/(x+3)>0`

Let `f(x)=((x-2)(x-3))/(x+3)`

Build the sign chart

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

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

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

`color(white)(aaaa)` `x-3` `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 (3,+oo)`

graph{(x^2-5x+6)/(x+3) [-148.4, 152, -74.4, 76]}

graph{(x^2-5x+6)/(x+3) [-0.486, 4.993, -1.132, 1.61]}