SOLUTION: solve the inequality algebraically: x^3-x^2-11x+3/2x^3-7x^2-19x+60 ≥ 0 I got (x+3)(x^2-4x+1)/(x+3)(2x^2-13x+20)≥0 so far. can someone please help me what I can do

Algebra ->  Inequalities -> SOLUTION: solve the inequality algebraically: x^3-x^2-11x+3/2x^3-7x^2-19x+60 ≥ 0 I got (x+3)(x^2-4x+1)/(x+3)(2x^2-13x+20)≥0 so far. can someone please help me what I can do       Log On


   

Question 1074611: solve the inequality algebraically: x^3-x^2-11x+3/2x^3-7x^2-19x+60 ≥ 0
I got (x+3)(x^2-4x+1)/(x+3)(2x^2-13x+20)≥0 so far.
can someone please help me what I can do from here?
Thank you.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Break up the number line into regions using the critical points (zeros) of the numerator and denominator.
Test a point in each region to determine if that region is part of the solution set.
Tedious but straightforward.
As an example,
%28%28x-1%29%28x%2B2%29%29%2F%28%28x-3%29%28x-5%29%29%3E=0
So the critical numbers are -2,1,3,5.
You'd break up the number line into regions,
Region 1 : (-infinity,-2)
Region 2 : (-2,1)
Region 3 : (1,3)
Region 4 : (3,5)
Region 5 : (5,infinity)
So for Region 1, you could use x=-3 to test the inequality.
The number you choose is not important just can't be the endpoints of the interval.
%28%28-3-1%29%28-3%2B2%29%29%2F%28%28-3-3%29%28-3-5%29%29%3E=0
%28%28-4%29%28-1%29%29%2F%28%28-6%29%28-8%29%29%3E=0
4%2F48%3E=0
True so Region 1 is part of the solution set.
You're more worried about the sign of the value and not the actual value.
Similarly,
Region 2, x=0
Region 3, x=2
Region 4, x=4
Region 5, x=6