SOLUTION: Solve the inequality {{{(x^4(3x-5)^2(x+1)^2)/(x^2-3x+2)>=0}}}

Algebra ->  Rational-functions -> SOLUTION: Solve the inequality {{{(x^4(3x-5)^2(x+1)^2)/(x^2-3x+2)>=0}}}      Log On


   



Question 1051852: Solve the inequality %28x%5E4%283x-5%29%5E2%28x%2B1%29%5E2%29%2F%28x%5E2-3x%2B2%29%3E=0
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%5E4%283x-5%29%5E2%28x%2B1%29%5E2%29%2F%28x%5E2-3x%2B2%29%3E=0

The denominator is the only thing not factored,
so factor it

%28x%5E4%283x-5%29%5E2%28x%2B1%29%5E2%29%2F%28%28x-1%29%28x-2%29%29%3E=0

The critical numbers are the zeros of all the factors:

0, 5%2F3, -1, 1, 2

Arrange them in numerical order:

-1, 0, 1, 5%2F3, 2

-1, 0, 1, 1%262%2F3, 2

and put them on a number line:

-----------o----o----o--o-o------
-3   -2   -1    0    1    2    3 

Choose test values less than the least, in between
them, and greater than the greatest:

Test values: -2, -0.5, 0.5, 1.5, 1.9, 3

If the test value gives %22%22%3E=0 when substituted
in the open interval which the test value is in is 
part of the solution; otherwise it's not.

Substituting -2 gives 161.333... which is %22%22%3E=0,
so %28matrix%281%2C3%2C-infinity%2C%22%2C%22%2C1%29%29 is part of
the solution.

Substituting -0.5 gives 0.176... which is %22%22%3E=0,
so %28matrix%281%2C3%2C-1%2C%22%2C%22%2C0%29%29 is part of
the solution.
  
Substituting 0.5 gives 2.296... which is %22%22%3E=0,
so %28matrix%281%2C3%2C0%2C%22%2C%22%2C1%29%29 is part of
the solution.
 
Substituting 1.5 gives -31.64... which is %22%22%3C0,
so %28matrix%281%2C3%2C1%2C%22%2C%22%2C5%2F3%29%29 is NOT part of
the solution.

Substituting 1.9 gives -596.71... which is %22%22%3C0,
so %28matrix%281%2C3%2C5%2F3%2C%22%2C%22%2C2%29%29 is NOT part of
the solution.

Substituting 3 gives 10368 which is %22%22%3E=0,
so %28matrix%281%2C3%2C1%2C%22%2C%22%2C2%29%29 is part of
the solution.

The critical numbers from the numerator 0,5/3, and -1
satisfy the given inequality since "equal to 0" is allowed.  
However the critical numbers from the denominator, 1 and 2, 
do not.

Solution set:



Edwin