Answer :
To solve the inequality
[tex]\[ (x-5)(x-2)(x+7) \geq 0, \][/tex]
we need to determine the values of \( x \) where the product of the expressions \((x-5)\), \((x-2)\), and \((x+7)\) is non-negative (i.e., either greater than or equal to zero).
### Step 1: Identify the critical points.
The critical points are the values of \( x \) that make each factor zero:
- \( x - 5 = 0 \) ⟹ \( x = 5 \)
- \( x - 2 = 0 \) ⟹ \( x = 2 \)
- \( x + 7 = 0 \) ⟹ \( x = -7 \)
The critical points are \( x = 5 \), \( x = 2 \), and \( x = -7 \).
### Step 2: Determine the intervals.
These critical points divide the number line into four intervals:
1. \((-\infty, -7)\)
2. \((-7, 2)\)
3. \((2, 5)\)
4. \((5, \infty)\)
### Step 3: Test the sign of the product in each interval.
Let's test a point from each interval to determine the sign of the expression in that interval.
1. Interval \((-\infty, -7)\):
- Choose a test point \( x = -8 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = -8 \) gives \((-8-5)(-8-2)(-8+7)\):
\(-13 \cdot -10 \cdot -1 = 130 \cdot -1 = -130\), which is negative.
2. Interval \((-7, 2)\):
- Choose a test point \( x = 0 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = 0 \) gives \((0-5)(0-2)(0+7)\):
\(-5 \cdot -2 \cdot 7 = 10 \cdot 7 = 70\), which is positive.
3. Interval \((2, 5)\):
- Choose a test point \( x = 3 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = 3 \) gives \((3-5)(3-2)(3+7)\):
\(-2 \cdot 1 \cdot 10 = -2 \cdot 10 = -20\), which is negative.
4. Interval \((5, \infty)\):
- Choose a test point \( x = 6 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = 6 \) gives \((6-5)(6-2)(6+7)\):
\(1 \cdot 4 \cdot 13 = 4 \cdot 13 = 52\), which is positive.
### Step 4: Include the critical points where the product is zero.
The expression equals zero at the critical points \( x = 5 \), \( x = 2 \), and \( x = -7 \). Since the inequality is \(\geq 0\), we include these points in our solution.
### Solution:
Combining the intervals where the product is non-negative and including the critical points, we have:
[tex]\[ x \in \left[ -7, 2 \right] \cup \left[ 5, \infty \right). \][/tex]
Thus, the solution to the inequality \((x-5)(x-2)(x+7) \geq 0\) is
[tex]\[ \boxed{\left[ -7, 2 \right] \cup \left[ 5, \infty \right)}. \][/tex]
[tex]\[ (x-5)(x-2)(x+7) \geq 0, \][/tex]
we need to determine the values of \( x \) where the product of the expressions \((x-5)\), \((x-2)\), and \((x+7)\) is non-negative (i.e., either greater than or equal to zero).
### Step 1: Identify the critical points.
The critical points are the values of \( x \) that make each factor zero:
- \( x - 5 = 0 \) ⟹ \( x = 5 \)
- \( x - 2 = 0 \) ⟹ \( x = 2 \)
- \( x + 7 = 0 \) ⟹ \( x = -7 \)
The critical points are \( x = 5 \), \( x = 2 \), and \( x = -7 \).
### Step 2: Determine the intervals.
These critical points divide the number line into four intervals:
1. \((-\infty, -7)\)
2. \((-7, 2)\)
3. \((2, 5)\)
4. \((5, \infty)\)
### Step 3: Test the sign of the product in each interval.
Let's test a point from each interval to determine the sign of the expression in that interval.
1. Interval \((-\infty, -7)\):
- Choose a test point \( x = -8 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = -8 \) gives \((-8-5)(-8-2)(-8+7)\):
\(-13 \cdot -10 \cdot -1 = 130 \cdot -1 = -130\), which is negative.
2. Interval \((-7, 2)\):
- Choose a test point \( x = 0 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = 0 \) gives \((0-5)(0-2)(0+7)\):
\(-5 \cdot -2 \cdot 7 = 10 \cdot 7 = 70\), which is positive.
3. Interval \((2, 5)\):
- Choose a test point \( x = 3 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = 3 \) gives \((3-5)(3-2)(3+7)\):
\(-2 \cdot 1 \cdot 10 = -2 \cdot 10 = -20\), which is negative.
4. Interval \((5, \infty)\):
- Choose a test point \( x = 6 \).
- Plugging in: \((x-5)(x-2)(x+7)\) at \( x = 6 \) gives \((6-5)(6-2)(6+7)\):
\(1 \cdot 4 \cdot 13 = 4 \cdot 13 = 52\), which is positive.
### Step 4: Include the critical points where the product is zero.
The expression equals zero at the critical points \( x = 5 \), \( x = 2 \), and \( x = -7 \). Since the inequality is \(\geq 0\), we include these points in our solution.
### Solution:
Combining the intervals where the product is non-negative and including the critical points, we have:
[tex]\[ x \in \left[ -7, 2 \right] \cup \left[ 5, \infty \right). \][/tex]
Thus, the solution to the inequality \((x-5)(x-2)(x+7) \geq 0\) is
[tex]\[ \boxed{\left[ -7, 2 \right] \cup \left[ 5, \infty \right)}. \][/tex]
