Question

How do you solve `(n^3-2n^2-n+2)/(n^3+3n^2+4n+12)<0`?

Answer 1

The solution is `n in (-3,-1) uu1,2)`

Explanation:

Factorise the numerator and the denominator

Start with the numerator

`n^3-2n^2-n+2=n^3-n-2n^2+2`

`=n(n^2-1)-2(n^2-1)`

`=(n^2-1)(n-2)`

`=(n+1)(n-1)(n-2)`

Proceed with the denominator

`n^3+3n^2++4n+12=n^3+4n+3n^2+12`

`=n(n^2+4)+3(n^2+4)`

`=(n^2+4)(n+3)`

Let `f(n)=(n^3-2n^2-n+2)/(n^3+3n^2++4n+12)=((n+1)(n-1)(n-2))/((n^2+4)(n+3))`

Build a sign chart

`color(white)(aaaa)` `n` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-3` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `1` `color(white)(aaaaa)` `2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `n+3` `color(white)(aaaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `n+1` `color(white)(aaaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaa)` `-` `color(white)(aaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `n-1` `color(white)(aaaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `n-2` `color(white)(aaaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(n)` `color(white)(aaaaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaa)` `-` `color(white)(aaa)` `+` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

Therefore,

`f(n)<0` when `n in (-3,-1) uu1,2)`

graph{(x^3-2x^2-x+2)/(x^3+4x+3x^2+12) [-12.66, 12.65, -6.33, 6.33]}