(X-1)(x+3)^2 (x-2)<0

To solve the inequality \((x - 1)(x + 3)^2 (x - 2) < 0\), we need to determine when the product is negative.

Step 1: Identify the critical points. To find the critical points, we set each factor to zero:

1. \(x - 1 = 0 \implies x = 1\)
2. \(x + 3 = 0 \implies x = -3\) (note: this factor is squared)
3. \(x - 2 = 0 \implies x = 2\)

The critical points are \(x = -3\), \(x = 1\), and \(x = 2\).

Step 2: Determine the intervals. The critical points divide the number line into four intervals:

1. \((- \infty, -3)\)
2. \((-3, 1)\)
3. \((1, 2)\)
4. \((2, \infty)\)

Step 3: Test each interval.

We will test a point from each interval to see if the product is less than zero.

1. For the interval \((- \infty, -3)\), choose \(x = -4\): \(( -4 - 1)( -4 + 3)^2( -4 - 2) = (-5)(1)(-6) = 30 > 0\)

2. For the interval \((-3, 1)\), choose \(x = 0\): \(( 0 - 1)( 0 + 3)^2( 0 - 2) = (-1)(9)(-2) = 18 > 0\)

3. For the interval \((1, 2)\), choose \(x = 1.5\): \(( 1.5 - 1)( 1.5 + 3)^2( 1.5 - 2) = (0.5)(20.25)(-0.5) = -5.0625 < 0\)

4. For the interval \((2, \infty)\), choose \(x = 3\): \(( 3 - 1)( 3 + 3)^2( 3 - 2) = (2)(36)(1) = 72 > 0\)

Step 4: Summary of test results.

• In the interval \((- \infty, -3)\): positive
• In the interval \((-3, 1)\): positive
• In the interval \((1, 2)\): negative (solution interval)
• In the interval \((2, \infty)\): positive

Step 5: Handle critical points.

• At \(x = -3\): the product is zero (since \((x + 3)^2 = 0\))
• At \(x = 1\): the product is zero (since \(x - 1 = 0\))
• At \(x = 2\): the product is zero (since \(x - 2 = 0\))

The inequality \(< 0\) excludes the points where the product equals zero.

Final Solution: The solution to the inequality \((x - 1)(x + 3)^2 (x - 2) < 0\) is:

\[ \boxed{(1, 2)} \]