The given inequality is:

$\frac{x^3 + x^2 - 2x - 2}{6x^2 + 7x + 2} \leq 0$

To solve this inequality, we will follow these steps:

Step 1: Factor the numerator and the denominator

We will try to factor both the numerator and the denominator if possible.

1. Numerator: $x^3 + x^2 - 2x - 2$

   We can group the terms and factor them: $x^2(x + 1) - 2(x + 1) = (x + 1)(x^2 - 2)$

   Therefore, the numerator becomes: $(x + 1)(x^2 - 2)$

2. Denominator: $6x^2 + 7x + 2$

   We factor this quadratic expression by finding two numbers that multiply to $6 \times 2 = 12$ and add to 7. The numbers are 3 and 4, so: $6x^2 + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1)$

   Therefore, the denominator becomes: $(3x + 2)(2x + 1)$

Step 2: Rewrite the inequality

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

Step 3: Determine critical points

The critical points are the values of $x$ where the expression is equal to zero or undefined. These occur when either the numerator is zero or the denominator is zero.

1. Numerator: $x + 1 = 0$ or $x^2 - 2 = 0$

   Solving these:

   • $x + 1 = 0 \Rightarrow x = -1$
   • $x^2 - 2 = 0 \Rightarrow x = \pm \sqrt{2}$

   Therefore, the roots of the numerator are $x = -1, \sqrt{2}, -\sqrt{2}$.

2. Denominator: $3x + 2 = 0$ or $2x + 1 = 0$

   Solving these:

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

   Therefore, the roots of the denominator are $x = -\frac{2}{3}, -\frac{1}{2}$.

Step 4: Test intervals

The critical points divide the number line into intervals. We will test a point from each interval to determine where the inequality holds.

The critical points are: $-\infty, -\sqrt{2}, -1, -\frac{2}{3}, -\frac{1}{2}, \sqrt{2}, \infty$

Test each interval for the sign of the expression.

1. Interval $(- \infty, -\sqrt{2})$: Choose $x = -2$
2. Interval $(- \sqrt{2}, -1)$: Choose $x = -1.5$
3. Interval $(-1, -\frac{2}{3})$: Choose $x = -0.9$
4. Interval $(- \frac{2}{3}, -\frac{1}{2})$: Choose $x = -0.6$
5. Interval $(- \frac{1}{2}, \sqrt{2})$: Choose $x = 0$
6. Interval $(\sqrt{2}, \infty)$: Choose $x = 2$

Step 5: Write the solution

After determining the sign of each interval, identify where the inequality is less than or equal to zero. Remember to include the points where the expression is equal to zero.

The solution will be the union of the intervals where the inequality holds.

If you need further clarification or a detailed solution for each interval, please let me know!