To solve the inequality:

$\frac{-2x^2 - 5x + 3}{x^2 - 4x - 5} \leq 0$

we'll go through the following steps:

1. Factor the numerator and denominator.
2. Find the zeros of the numerator and denominator.
3. Determine the intervals of sign changes.
4. Test intervals and apply inequality conditions.

Step 1: Factor the Numerator and Denominator

Factor the Numerator: $-2x^2 - 5x + 3$

Using factoring methods, we get: $-2x^2 - 5x + 3 = -(2x + 3)(x - 1)$

Factor the Denominator: $x^2 - 4x - 5$

This factors as: $x^2 - 4x - 5 = (x - 5)(x + 1)$

So we can rewrite the inequality as: $\frac{-(2x + 3)(x - 1)}{(x - 5)(x + 1)} \leq 0$

Step 2: Find the Zeros and Undefined Points

The inequality will change sign at the zeros and undefined points of the expression. Set each factor to zero:

• $2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$
• $x - 1 = 0 \Rightarrow x = 1$
• $x - 5 = 0 \Rightarrow x = 5$
• $x + 1 = 0 \Rightarrow x = -1$

These values divide the number line into intervals, and the points where the expression is undefined are $x = 5$ and $x = -1$.

Step 3: Test Intervals and Apply Inequality Condition

The intervals created by the critical points are: $(-\infty, -1), \; (-1, -\frac{3}{2}), \; (-\frac{3}{2}, 1), \; (1, 5), \; (5, \infty)$

We'll test the sign of $\frac{-(2x + 3)(x - 1)}{(x - 5)(x + 1)}$ in each interval:

1. For $x \in (-\infty, -1)$: All terms are negative, so the expression is positive.
2. For $x \in (-1, -\frac{3}{2})$: Choose $x = -1.5$. The expression is negative.
3. For $x \in (-\frac{3}{2}, 1)$: Choose $x = 0$. The expression is positive.
4. For $x \in (1, 5)$: Choose $x = 2$. The expression is negative.
5. For $x \in (5, \infty)$: All terms are positive, so the expression is positive.

Step 4: Satisfy the Inequality Condition $\leq 0$

We are looking for intervals where the expression is less than or equal to zero.

• From our sign analysis, the intervals where $\frac{-(2x + 3)(x - 1)}{(x - 5)(x + 1)} \leq 0$ are:
  • $(-1, -\frac{3}{2}]$
  • $[1, 5)$

Final Answer

The solution set for the inequality is: $x \in \left(-\frac{3}{2}, -1\right) \cup [1, 5)$

Do you have any questions, or would you like more details on a specific step?

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Related Questions:

1. How do we determine the sign of a rational expression in each interval?
2. What is the impact of undefined points on inequalities involving rational expressions?
3. How does flipping signs affect inequality solutions in expressions with negative factors?
4. What other methods can be used to solve rational inequalities?
5. Why is it important to consider intervals carefully in inequalities?

Tip: Remember, always exclude points where the expression is undefined in the final answer!