Answer :
To solve the inequality \((x+2)(x-3)(x+8) \leq 0\), let's go through a step-by-step process:
1. Identify the critical points:
The inequality changes its behavior at the values of \(x\) where the expression equals zero. Solve for the roots of the equation:
[tex]\[ (x+2)(x-3)(x+8) = 0 \][/tex]
The critical points are:
[tex]\[ x = -2, x = 3, x = -8 \][/tex]
2. Determine the intervals:
These critical points divide the number line into four intervals:
[tex]\[ (-\infty, -8), (-8, -2), (-2, 3), (3, \infty) \][/tex]
3. Test the sign of the expression in each interval:
Choose test points from each interval:
- For \((- \infty, -8)\), use \(x = -9\):
[tex]\[ (-9+2)(-9-3)(-9+8) = (-7)(-12)(-1) = -84 \quad \text{(negative)} \][/tex]
- For \((-8, -2)\), use \(x = -5\):
[tex]\[ (-5+2)(-5-3)(-5+8) = (-3)(-8)(3) = 72 \quad \text{(positive)} \][/tex]
- For \((-2, 3)\), use \(x = 0\):
[tex]\[ (0+2)(0-3)(0+8) = (2)(-3)(8) = -48 \quad \text{(negative)} \][/tex]
- For \((3, \infty)\), use \(x = 4\):
[tex]\[ (4+2)(4-3)(4+8) = (6)(1)(12) = 72 \quad \text{(positive)} \][/tex]
4. Include the endpoints where the expression is zero:
Since the inequality is \(\leq 0\), include the points where the expression is exactly zero:
[tex]\[ x = -8, x = -2, x = 3 \][/tex]
5. Combine the intervals where the expression is negative or zero:
The expression \((x+2)(x-3)(x+8) \leq 0\) holds in the intervals where we found either a negative value or zero:
[tex]\[ (-\infty, -8] \cup [-2, 3] \][/tex]
So, the solution to the inequality \((x+2)(x-3)(x+8) \leq 0\) is:
[tex]\[ \boxed{(-\infty, -8] \cup [-2, 3]} \][/tex]
1. Identify the critical points:
The inequality changes its behavior at the values of \(x\) where the expression equals zero. Solve for the roots of the equation:
[tex]\[ (x+2)(x-3)(x+8) = 0 \][/tex]
The critical points are:
[tex]\[ x = -2, x = 3, x = -8 \][/tex]
2. Determine the intervals:
These critical points divide the number line into four intervals:
[tex]\[ (-\infty, -8), (-8, -2), (-2, 3), (3, \infty) \][/tex]
3. Test the sign of the expression in each interval:
Choose test points from each interval:
- For \((- \infty, -8)\), use \(x = -9\):
[tex]\[ (-9+2)(-9-3)(-9+8) = (-7)(-12)(-1) = -84 \quad \text{(negative)} \][/tex]
- For \((-8, -2)\), use \(x = -5\):
[tex]\[ (-5+2)(-5-3)(-5+8) = (-3)(-8)(3) = 72 \quad \text{(positive)} \][/tex]
- For \((-2, 3)\), use \(x = 0\):
[tex]\[ (0+2)(0-3)(0+8) = (2)(-3)(8) = -48 \quad \text{(negative)} \][/tex]
- For \((3, \infty)\), use \(x = 4\):
[tex]\[ (4+2)(4-3)(4+8) = (6)(1)(12) = 72 \quad \text{(positive)} \][/tex]
4. Include the endpoints where the expression is zero:
Since the inequality is \(\leq 0\), include the points where the expression is exactly zero:
[tex]\[ x = -8, x = -2, x = 3 \][/tex]
5. Combine the intervals where the expression is negative or zero:
The expression \((x+2)(x-3)(x+8) \leq 0\) holds in the intervals where we found either a negative value or zero:
[tex]\[ (-\infty, -8] \cup [-2, 3] \][/tex]
So, the solution to the inequality \((x+2)(x-3)(x+8) \leq 0\) is:
[tex]\[ \boxed{(-\infty, -8] \cup [-2, 3]} \][/tex]
