The inequality given is:

$(x - 7)(x + 1) > 0$

To solve this, we'll break it down step by step.

Step 1: Find the critical points

The critical points occur where the product equals zero. That is, solve:

$(x - 7)(x + 1) = 0$

So, the critical points are: $x = 7 \quad \text{and} \quad x = -1$

Step 2: Determine intervals on the number line

The critical points divide the number line into three intervals:

1. $(-\infty, -1)$
2. $(-1, 7)$
3. $(7, \infty)$

Step 3: Test points in each interval

We need to determine where the product $(x - 7)(x + 1)$ is positive. Let's test points in each interval.

1. For $x = -2$ (in the interval $(-\infty, -1)$): $(x - 7)(x + 1) = (-2 - 7)(-2 + 1) = (-9)(-1) = 9 > 0$ So, the inequality is true in this interval.

2. For $x = 0$ (in the interval $(-1, 7)$): $(x - 7)(x + 1) = (0 - 7)(0 + 1) = (-7)(1) = -7 < 0$ So, the inequality is false in this interval.

3. For $x = 8$ (in the interval $(7, \infty)$): $(x - 7)(x + 1) = (8 - 7)(8 + 1) = (1)(9) = 9 > 0$ So, the inequality is true in this interval.

Step 4: Analyze the critical points

At $x = -1$ and $x = 7$, the product is zero: $(x - 7)(x + 1) = 0$ Since the inequality is strict ( $> 0$ ), we exclude these points from the solution.

Step 5: Conclusion

The solution to the inequality is:

$x \in (-\infty, -1) \cup (7, \infty)$

Graph of the solution on the number line:

• A solid line in the intervals $(-\infty, -1)$ and $(7, \infty)$, excluding the points $x = -1$ and $x = 7$.
• Open circles at $x = -1$ and $x = 7$ to indicate that these points are not included in the solution.

Would you like a visual graph of the solution?

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Questions:

1. How would the solution change if the inequality were $\geq 0$?
2. What happens if the critical points are included in the solution?
3. Can you apply the same method to inequalities involving cubic terms?
4. How do open and closed intervals affect the graphical representation of inequalities?
5. How can you solve inequalities with more than two factors?

Tip: Always check the sign of each interval when solving inequalities by testing a point within each interval.