Answer :
To solve the inequality [tex]\((x+6)^2 (x+3) (x-1) > 0\)[/tex], we need to determine the intervals where this expression is positive. Here are the steps to find the solution:
1. Identify the critical points:
These are the values of [tex]\(x\)[/tex] that make the expression equal to zero. The critical points can be found by setting each factor of the expression equal to zero:
- [tex]\((x+6)^2 = 0\)[/tex]
- [tex]\(x+3 = 0\)[/tex]
- [tex]\(x-1 = 0\)[/tex]
Solving these equations gives the critical points:
[tex]\[ x = -6, \quad x = -3, \quad x = 1 \][/tex]
2. Determine the sign of the expression in the intervals:
The critical points divide the number line into several intervals. We need to test the sign of the expression in each interval. The intervals are:
[tex]\[ (-\infty, -6), \quad (-6, -3), \quad (-3, 1), \quad (1, +\infty) \][/tex]
3. Test a point in each interval:
Choose a test point from each interval to determine whether the expression is positive or negative within that interval.
- Interval [tex]\((- \infty, -6)\)[/tex]:
Choose [tex]\(x = -7\)[/tex]:
[tex]\[ ((-7+6)^2) \cdot (-7+3) \cdot (-7-1) = 1^2 \cdot (-4) \cdot (-8) = 32 \quad (\text{positive}) \][/tex]
- Interval [tex]\((-6, -3)\)[/tex]:
Choose [tex]\(x = -5\)[/tex]:
[tex]\[ ((-5+6)^2) \cdot (-5+3) \cdot (-5-1) = 1^2 \cdot (-2) \cdot (-6) = 12 \quad (\text{positive}) \][/tex]
- Interval [tex]\((-3, 1)\)[/tex]:
Choose [tex]\(x = 0\)[/tex]:
[tex]\[ ((0+6)^2) \cdot (0+3) \cdot (0-1) = 36 \cdot 3 \cdot (-1) = -108 \quad (\text{negative}) \][/tex]
- Interval [tex]\((1, \infty)\)[/tex]:
Choose [tex]\(x = 2\)[/tex]:
[tex]\[ ((2+6)^2) \cdot (2+3) \cdot (2-1) = 64 \cdot 5 \cdot 1 = 320 \quad (\text{positive}) \][/tex]
4. Combine the intervals where the inequality holds:
The expression is positive in the intervals where our test points resulted in positive values.
Therefore, the solution to the inequality [tex]\((x+6)^2 (x+3) (x-1) > 0\)[/tex] is:
[tex]\[ (-\infty, -6) \cup (-6, -3) \cup (1, \infty) \][/tex]
This is the answer in interval notation.
Thus, the inequality [tex]\((x+6)^2 (x+3) (x-1) > 0\)[/tex] holds for [tex]\(x\)[/tex] in the intervals
\[
(-\infty, -6) \cup (-6, -3) \cup (1, \infty).
1. Identify the critical points:
These are the values of [tex]\(x\)[/tex] that make the expression equal to zero. The critical points can be found by setting each factor of the expression equal to zero:
- [tex]\((x+6)^2 = 0\)[/tex]
- [tex]\(x+3 = 0\)[/tex]
- [tex]\(x-1 = 0\)[/tex]
Solving these equations gives the critical points:
[tex]\[ x = -6, \quad x = -3, \quad x = 1 \][/tex]
2. Determine the sign of the expression in the intervals:
The critical points divide the number line into several intervals. We need to test the sign of the expression in each interval. The intervals are:
[tex]\[ (-\infty, -6), \quad (-6, -3), \quad (-3, 1), \quad (1, +\infty) \][/tex]
3. Test a point in each interval:
Choose a test point from each interval to determine whether the expression is positive or negative within that interval.
- Interval [tex]\((- \infty, -6)\)[/tex]:
Choose [tex]\(x = -7\)[/tex]:
[tex]\[ ((-7+6)^2) \cdot (-7+3) \cdot (-7-1) = 1^2 \cdot (-4) \cdot (-8) = 32 \quad (\text{positive}) \][/tex]
- Interval [tex]\((-6, -3)\)[/tex]:
Choose [tex]\(x = -5\)[/tex]:
[tex]\[ ((-5+6)^2) \cdot (-5+3) \cdot (-5-1) = 1^2 \cdot (-2) \cdot (-6) = 12 \quad (\text{positive}) \][/tex]
- Interval [tex]\((-3, 1)\)[/tex]:
Choose [tex]\(x = 0\)[/tex]:
[tex]\[ ((0+6)^2) \cdot (0+3) \cdot (0-1) = 36 \cdot 3 \cdot (-1) = -108 \quad (\text{negative}) \][/tex]
- Interval [tex]\((1, \infty)\)[/tex]:
Choose [tex]\(x = 2\)[/tex]:
[tex]\[ ((2+6)^2) \cdot (2+3) \cdot (2-1) = 64 \cdot 5 \cdot 1 = 320 \quad (\text{positive}) \][/tex]
4. Combine the intervals where the inequality holds:
The expression is positive in the intervals where our test points resulted in positive values.
Therefore, the solution to the inequality [tex]\((x+6)^2 (x+3) (x-1) > 0\)[/tex] is:
[tex]\[ (-\infty, -6) \cup (-6, -3) \cup (1, \infty) \][/tex]
This is the answer in interval notation.
Thus, the inequality [tex]\((x+6)^2 (x+3) (x-1) > 0\)[/tex] holds for [tex]\(x\)[/tex] in the intervals
\[
(-\infty, -6) \cup (-6, -3) \cup (1, \infty).
