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We are given the inequality \( x^3 - 5x^2 - x + 5 > 0 \) to solve and represent using interval notation and on a number line.

### Steps to solve:

1. **Factor the cubic expression**: Start by factoring \( x^3 - 5x^2 - x + 5 \).

   Using synthetic division or factoring methods, the expression factors as:

   \[
   (x - 5)(x^2 - 1) = 0
   \]

   Then factor \( x^2 - 1 \) further:

   \[
   (x - 5)(x - 1)(x + 1) = 0
   \]

2. **Find the critical points**: Set each factor equal to zero to find the roots (critical points):

   \[
   x - 5 = 0 \implies x = 5
   \]
   \[
   x - 1 = 0 \implies x = 1
   \]
   \[
   x + 1 = 0 \implies x = -1
   \]

   So, the critical points are \( x = -1 \), \( x = 1 \), and \( x = 5 \).

3. **Test intervals**: The critical points divide the number line into four intervals: \( (-\infty, -1) \), \( (-1, 1) \), \( (1, 5) \), and \( (5, \infty) \). Test a value from each interval in the inequality \( (x - 5)(x - 1)(x + 1) > 0 \).

   - For \( x \in (-\infty, -1) \), say \( x = -2 \): 
     \[
     (-2 - 5)(-2 - 1)(-2 + 1) = (-7)(-3)(-1) = -21 < 0
     \]
     So, no solution in this interval.
   
   - For \( x \in (-1, 1) \), say \( x = 0 \):
     \[
     (0 - 5)(0 - 1)(0 + 1) = (-5)(-1)(1) = 5 > 0
     \]
     So, this interval is part of the solution.

   - For \( x \in (1, 5) \), say \( x = 2 \):
     \[
     (2 - 5)(2 - 1)(2 + 1) = (-3)(1)(3) = -9 < 0
     \]
     No solution here.

   - For \( x \in (5, \infty) \), say \( x = 6 \):
     \[
     (6 - 5)(6 - 1)(6 + 1) = (1)(5)(7) = 35 > 0
     \]
     So, this interval is part of the solution.

4. **Conclusion**: The inequality is satisfied on the intervals \( (-1, 1) \) and \( (5, \infty) \).

   Thus, the solution in interval notation is:
   \[
   (-1, 1) \cup (5, \infty)
   \]

### Graph the solution set:

You would plot the solution on the number line, with open circles at \( -1 \), \( 1 \), and \( 5 \), and shading the intervals \( (-1, 1) \) and \( (5, \infty) \).

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Would you like any further details or explanations?

Here are some related questions to deepen your understanding:

1. How do we perform synthetic division to factor cubic polynomials?
2. What happens when we test points in different intervals for polynomial inequalities?
3. How can we confirm that an inequality's solution excludes the critical points?
4. What are the implications of testing boundary points for strict inequalities?
5. How do graphs of cubic functions typically behave at their critical points?

**Tip:** Always double-check each test point you use in different intervals to avoid sign mistakes when solving inequalities.

Solve the inequality x^3 - 5x^2 - x + 5 > 0 and represent the solution using interval notation and graph the solution set.

This solution guides you through solving the cubic inequality x^3 - 5x^2 - x + 5 > 0, using factorization and interval testing. The solution is represented in interval notation and includes graphing the solution on a number line. Suitable for students in Grades 10-12.

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