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

1 Answer
Mar 2, 2018

Answer:

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]}