Question

(-x-7)(-x-1)(x-3)(x-2)>=0

Answer 1

-х-7>0
х=-7
-х-1>0
х=-1
х-3>0
х>3
х-2>0
х=2

Answer 2

......................

Answer 3

To solve the inequality (-x-7)(-x-1)(x-3)(x-2) >= 0, we can use the concept of intervals and the sign of the expression within each interval. Let's break it down step by step:

1. Find the critical points: To determine the intervals, we need to find the critical points where the expression becomes zero. These occur when each factor is equal to zero. So we solve the following equations:

-x - 7 = 0 => x = -7 -x - 1 = 0 => x = -1 x - 3 = 0 => x = 3 x - 2 = 0 => x = 2

2. Create intervals: We have four critical points: -7, -1, 2, and 3. These points divide the real number line into five intervals:

Interval 1: (-∞, -7) Interval 2: (-7, -1) Interval 3: (-1, 2) Interval 4: (2, 3) Interval 5: (3, +∞)

3. Test intervals: Now we need to determine the sign of the expression (-x-7)(-x-1)(x-3)(x-2) within each interval. We can choose test points within each interval and evaluate the expression.

Interval 1: (-∞, -7) Choose x = -8: (-(-8)-7)(-(-8)-1)(-8-3)(-8-2) = (-1)(-7)(-11)(-10) = -770 The expression is negative in this interval.

Interval 2: (-7, -1) Choose x = 0: (-(0)-7)(-(0)-1)(0-3)(0-2) = (-7)(-1)(-3)(-2) = -42 The expression is negative in this interval.

Interval 3: (-1, 2) Choose x = 1: (-(1)-7)(-(1)-1)(1-3)(1-2) = (-8)(-2)(-2)(-1) = 32 The expression is positive in this interval.

Interval 4: (2, 3) Choose x = 2.5: (-(2.5)-7)(-(2.5)-1)(2.5-3)(2.5-2) = (-9.5)(-3.5)(-0.5)(0.5) = -6.125 The expression is negative in this interval.

Interval 5: (3, +∞) Choose x = 4: (-(4)-7)(-(4)-1)(4-3)(4-2) = (-11)(-5)(1)(2) = -110 The expression is negative in this interval.

4. Determine the solution: Now, we can summarize the sign of the expression within each interval:

Interval 1: (-∞, -7) -> Negative Interval 2: (-7, -1) -> Negative Interval 3: (-1, 2) -> Positive Interval 4: (2, 3) -> Negative Interval 5: (3, +∞) -> Negative

The inequality (-x-7)(-x-1)(x-3)(x-2) >= 0 holds true when the expression is either zero or positive. Based on the sign analysis above, the solution is:

x ∈ (-1, 2] U (3, +∞)