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)ncolor(white)(aaa)-oocolor(white)(aaa)-3/2color(white)(aaa)-4/3color(white)(aaaa)-1/2color(white)(aaaa)1/2color(white)(aaaa)+oo
color(white)(aaaa)2n-3color(white)(aaaa)-color(white)(a)||color(white)(aa)+color(white)(aaaa)+color(white)(aaaa)+color(white)(aaaa)+
color(white)(aaaa)3n+4color(white)(aaaa)-color(white)(a)||color(white)(aa)-color(white)(aaaa)+color(white)(aaaa)+color(white)(aaaa)+
color(white)(aaaa)2n+1color(white)(aaaa)-color(white)(a)||color(white)(aa)-color(white)(aaaa)-color(white)(aaaa)+color(white)(aaaa)+
color(white)(aaaa)2n-1color(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]}
