(х-2)(х+3)(х-1)≥0

To solve the inequality (х-2)(х+3)(х-1)≥0, we need to find the intervals where the expression is greater than or equal to zero.

First, let's find the critical points where the expression becomes zero: (x - 2) = 0 ---> x = 2 (x + 3) = 0 ---> x = -3 (x - 1) = 0 ---> x = 1

Next, we can plot these critical points on a number line to determine the intervals where the expression is positive or negative:

-3 1 2

Now, let's test a value in each interval: For x < -3, let's substitute x = -4: (-4 - 2)(-4 + 3)(-4 - 1) = (-6)(-1)(-5) = 30 Since this is positive (greater than zero), the expression is positive in this interval.

For -3 < x < 1, let's substitute x = 0: (0 - 2)(0 + 3)(0 - 1) = (-2)(3)(-1) = 6 Since this is positive (greater than zero), the expression is positive in this interval.

For 1 < x < 2, let's substitute x = 1.5 (a value between 1 and 2): (1.5 - 2)(1.5 + 3)(1.5 - 1) = (-0.5)(4.5)(0.5) = -0.5625 Since this is negative (less than zero), the expression is negative in this interval.

For x > 2, let's substitute x = 3: (3 - 2)(3 + 3)(3 - 1) = (1)(6)(2) = 12 Since this is positive (greater than zero), the expression is positive in this interval.

Now, we can summarize the intervals where the expression (х-2)(х+3)(х-1) is greater than or equal to zero: (-∞, -3] U [1, 2] U [2, +∞)

So, the solution to the inequality (х-2)(х+3)(х-1)≥0 is (-∞, -3] U [1, 2] U [2, +∞).