# How do you solve #(n^3+3n^2-4n-12)/(n^3-5n^2+4n-20)<=0#?

##### 1 Answer

#### Explanation:

#n^3+3n^2-4n-12 = (n^3+3n^2)-(4n+12)#

#color(white)(n^3+3n^2-4n-12) = n^2(n+3)-4(n+3)#

#color(white)(n^3+3n^2-4n-12) = (n^2-4)(n+3)#

#color(white)(n^3+3n^2-4n-12) = (n^2-2^2)(n+3)#

#color(white)(n^3+3n^2-4n-12) = (n-2)(n+2)(n+3)#

#n^3-5n^2+4n-20 = (n^3-5n^2)+(4n-20)#

#color(white)(n^3-5n^2+4n-20) = n^2(n-5)+4(n-5)#

#color(white)(n^3-5n^2+4n-20) = (n^2+4)(n-5)#

So the numerator and denominator have no common factors and any linear factors occur precisely once.

The zeros of one or other are at

So the quotient changes sign at each of these points.

Note that for large positive values of

Hence the sign of the quotient in each of the following intervals is as follows:

#(5, oo): +#

#(2, 5): -#

#(-2, 2): +#

#(-3, -2): -#

#(-oo, -3): +#

Hence the inequality is satisfied for:

#n in [-3, -2] uu [2, 5)#

graph{ (x^3+3x^2-4x-12)/(x^3-5x^2+4x-20) [-8.085, 11.915, -4.76, 5.24]}