Question

How do solve `x^3+2x^2<=8x` and write the answer as a inequality and interval notation?

Answer 1

The answer is `-oo<=x<=-4` and `0<=x<=2`

Explanation:

Let us factorise the inequality and make a sign chart

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

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

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

So the values are `x=0`, `x=2`and `x=-4`

`x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `-4` `color(white)(aaaaa)` `0` `color(white)(aaaaa)` `2` `color(white)(aaaaa)` `+oo`
`x` `color(white)(aaaaa)` `color(white)(aaaa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`
`x-2` `color(white)(aaaa)` `color(white)(aa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`
`x+4` `color(white)(aaaa)` `color(white)(aa)` `-` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`
`X` `color(white)(aaaaaa)` `color(white)(aa)` `-` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`
So the answer is `-oo<=x<=-4` and `0<=x<=2`