To find the intervals where the expression is greater than 0, we need to determine the sign of the expression in each interval and use the properties of inequality.
First, let's find the critical points by setting each factor equal to zero and solving for x:
(2x+1) = 0 2x = -1 x = -1/2
(1-2x) = 0 -2x = -1 x = 1/2
(x-1) = 0 x = 1
(2-3x) = 0 -3x = -2 x = 2/3
Now, let's test each interval:
Interval 1: (-∞, -1/2) Pick x = -1: (-)(+)(-)(+) > 0 is true
Interval 2: (-1/2, 1/2) Pick x = 0: (+)(+)(-)(+) > 0 is false
Interval 3: (1/2, 1) Pick x = 1: (+)(-)(+)(+) > 0 is true
Interval 4: (1, 2/3) Pick x = 3/4: (+)(-)(+)(-) > 0 is false
Interval 5: (2/3, ∞) Pick x = 1: (+)(-)(+)(-) > 0 is true
Therefore, the solution to the inequality (2x+1)(1-2x)(x-1)(2-3x) > 0 is x ∈ (-∞, -1/2) U (1/2, 1) U (2/3, ∞).
To find the intervals where the expression is greater than 0, we need to determine the sign of the expression in each interval and use the properties of inequality.
First, let's find the critical points by setting each factor equal to zero and solving for x:
(2x+1) = 0
2x = -1
x = -1/2
(1-2x) = 0
-2x = -1
x = 1/2
(x-1) = 0
x = 1
(2-3x) = 0
-3x = -2
x = 2/3
Now, let's test each interval:
Interval 1: (-∞, -1/2)
Pick x = -1:
(-)(+)(-)(+) > 0 is true
Interval 2: (-1/2, 1/2)
Pick x = 0:
(+)(+)(-)(+) > 0 is false
Interval 3: (1/2, 1)
Pick x = 1:
(+)(-)(+)(+) > 0 is true
Interval 4: (1, 2/3)
Pick x = 3/4:
(+)(-)(+)(-) > 0 is false
Interval 5: (2/3, ∞)
Pick x = 1:
(+)(-)(+)(-) > 0 is true
Therefore, the solution to the inequality (2x+1)(1-2x)(x-1)(2-3x) > 0 is x ∈ (-∞, -1/2) U (1/2, 1) U (2/3, ∞).