Question

X(x+3)(x-2)(x+4)(x-6)≤0

Answer 1

Смотри....................

Answer 2

To solve the inequality $X(x+3)(x-2)(x+4)(x-6) \leq 0$, we need to find the intervals where the expression is less than or equal to zero.

To do this, we'll use the concept of intervals and test the sign of the expression within each interval. The expression will change sign whenever any of the factors $X$, $x+3$, $x-2$, $x+4$, or $x-6$ changes sign.

Let's break it down step by step:

Step 1: Find critical points (points where the expression changes sign):

1. $X = 0$ (from the factor $X$)
2. $x + 3 = 0 \Rightarrow x = -3$ (from the factor $x + 3$)
3. $x - 2 = 0 \Rightarrow x = 2$ (from the factor $x - 2$)
4. $x + 4 = 0 \Rightarrow x = -4$ (from the factor $x + 4$)
5. $x - 6 = 0 \Rightarrow x = 6$ (from the factor $x - 6$)

Step 2: Create intervals and test the sign within each interval:

Choose a value in each interval and evaluate the expression to determine its sign.

Interval 1: $-\infty < x < -4$ (Test $x = -5$) $X(-5+3)(-5-2)(-5+4)(-5-6) = X(-2)(-7)(-1)(-11) = 154X$

Interval 2: $-4 < x < -3$ (Test $x = -3.5$) $X(-3.5+3)(-3.5-2)(-3.5+4)(-3.5-6) = X(-0.5)(-5.5)(0.5)(-9.5) = 13.75X$

Interval 3: $-3 < x < 2$ (Test $x = 0$) $X(0+3)(0-2)(0+4)(0-6) = X(3)(-2)(4)(-6) = -144X$

Interval 4: $2 < x < 6$ (Test $x = 3$) $X(3+3)(3-2)(3+4)(3-6) = X(6)(1)(7)(-3) = -126X$

Interval 5: $x > 6$ (Test $x = 7$) $X(7+3)(7-2)(7+4)(7-6) = X(10)(5)(11)(1) = 550X$

Step 3: Determine the intervals where $X(x+3)(x-2)(x+4)(x-6) \leq 0$:

For the expression to be less than or equal to zero, we need to find the intervals where the sign of the expression is negative (or zero). These are the intervals:

1. $-\infty < x \leq -4$
2. $-3 \leq x \leq 2$

Step 4: Write the final solution:

The solution to the inequality is $X(x+3)(x-2)(x+4)(x-6) \leq 0$ for $x$ in the intervals $-\infty < x \leq -4$ and $-3 \leq x \leq 2$.