Question

How do you solve `(12n^3+16n^2-3n-4)/(8n^3+12n^2+10n+15)>0`?

Answer 1

The solution is `n in (-oo,-3/2)uu(-4/3,-1/2)uu(1/2,+oo)`

Explanation:

Start by factorising the numerator and the denominator

`12n^3+16n^2-3n-4=12n^3-3n+16n^2-4`

`=3n(4n^2-1)+4(4n^2-1)`

`=(4n^2-1)(3n+4)`

`=(2n+1)(2n-1)(3n+4)`

`8n^3+12n^2+10n+15=8n^3+10n+12n^2+15`

`=2n(4n^2+5)+3(4n^2+5)`

`=(4n^2+5)(2n+3)`

Finally,
`(12n^3+16n^2-3n-4)/(8n^3+12n^2+10n+15)=((2n+1)(2n-1)(3n+4))/((4n^2+5)(2n+3))`

Let `f(n)=((2n+1)(2n-1)(3n+4))/((4n^2+5)(2n+3))`

The term `(4n^2+5)>0`

We can construct the sign chart

`color(white)(aaaa)` `n` `color(white)(aaa)` `-oo` `color(white)(aaa)` `-3/2` `color(white)(aaa)` `-4/3` `color(white)(aaaa)` `-1/2` `color(white)(aaaa)` `1/2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `2n-3` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `3n+4` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2n+1` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2n-1` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

`f(n)>0` when `n in (-oo,-3/2)uu(-4/3,-1/2)uu(1/2,+oo)`

graph{(12x^3+16x^2-3x-4)/(8x^3+12x^2+10x+15) [-5.944, 3.92, -2.265, 2.665]}