Question

How do solve `2x^3+x^2>6x` and write the answer as a inequality and interval notation?

Answer 1

The solutions are `-2 < x < 0` and `x > 3/2`
Or in interval notation, `x in ]-2,0[uu]3/2,+oo[`

Explanation:

Let's rewrite and factorise the inequality

`2x^3+x^2>6x`

`2x^3+x^2-6x>0`

`x(2x^2+x-6)>0`

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

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

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaaa)` `0` `color(white)(aaaaa)` `3/2` `color(white)(aaaaaa)` `+oo`

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

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

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

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

Therefore,

`f(x)>0` when `-2 < x < 0` and `x > 3/2`

Or in interval notation, `x in ]-2,0[uu]3/2,+oo[`