Question

How do you solve `x^3+2x^2<=8x` using a sign chart?

Answer 1

The solution is `x in ]-oo,-4]uu[0, 2]`

Explanation:

Let's rewrite and factorise the inequality

`x^3+2x^2<=8x`

`x^3+2x^2-8x<=0`

`x(x^2+2x-8)<=0`

`x(x-2)(x+4)<=0`

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

Now, we can build the sign chart

`color(white)(aaaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaa)` `0` `color(white)(aaaaa)` `2` `color(white)(aaaaa)` `+oo`

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

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

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

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

Therefore,

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