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

1 Answer
Nov 11, 2017

Answer:

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]}