We are tasked with solving the equation:

\[
\frac{4}{a^2 - a - 2} - \frac{2}{a^2 + 3a + 2} = \frac{-3}{4 - a^2}
\]

### Step 1: Factor the quadratic expressions
We'll begin by factoring the quadratic expressions in the denominators:

- \( a^2 - a - 2 \) factors as:
  \[
  a^2 - a - 2 = (a - 2)(a + 1)
  \]

- \( a^2 + 3a + 2 \) factors as:
  \[
  a^2 + 3a + 2 = (a + 1)(a + 2)
  \]

- \( 4 - a^2 \) can be written as a difference of squares:
  \[
  4 - a^2 = (2 - a)(2 + a)
  \]

Now the equation becomes:

\[
\frac{4}{(a - 2)(a + 1)} - \frac{2}{(a + 1)(a + 2)} = \frac{-3}{(2 - a)(2 + a)}
\]

### Step 2: Simplify the equation
Notice that \( 2 - a = -(a - 2) \). So we can rewrite the right-hand side as:

\[
\frac{-3}{(2 - a)(2 + a)} = \frac{3}{(a - 2)(a + 2)}
\]

Thus, the equation becomes:

\[
\frac{4}{(a - 2)(a + 1)} - \frac{2}{(a + 1)(a + 2)} = \frac{3}{(a - 2)(a + 2)}
\]

### Step 3: Combine the left-hand side
The common denominator on the left-hand side is \( (a - 2)(a + 1)(a + 2) \). We'll rewrite both fractions with this common denominator:

\[
\frac{4(a + 2)}{(a - 2)(a + 1)(a + 2)} - \frac{2(a - 2)}{(a - 2)(a + 1)(a + 2)}
\]

Now simplify the numerators:

\[
\frac{4(a + 2) - 2(a - 2)}{(a - 2)(a + 1)(a + 2)}
\]

Distribute in the numerators:

\[
4(a + 2) = 4a + 8
\]
\[
2(a - 2) = 2a - 4
\]

Thus, the equation becomes:

\[
\frac{(4a + 8) - (2a - 4)}{(a - 2)(a + 1)(a + 2)} = \frac{3}{(a - 2)(a + 2)}
\]

Simplifying the numerator on the left-hand side:

\[
(4a + 8) - (2a - 4) = 4a + 8 - 2a + 4 = 2a + 12
\]

So, the equation is now:

\[
\frac{2a + 12}{(a - 2)(a + 1)(a + 2)} = \frac{3}{(a - 2)(a + 2)}
\]

### Step 4: Eliminate the denominators
Since both sides of the equation have a common factor of \( (a - 2)(a + 2) \), we can multiply through by \( (a - 2)(a + 2) \) to eliminate the denominators:

\[
\frac{2a + 12}{a + 1} = 3
\]

### Step 5: Solve for \( a \)
Multiply both sides by \( a + 1 \) to get rid of the denominator on the left-hand side:

\[
2a + 12 = 3(a + 1)
\]

Expand the right-hand side:

\[
2a + 12 = 3a + 3
\]

Now, solve for \( a \):

\[
12 - 3 = 3a - 2a
\]

\[
9 = a
\]

Thus, the solution is \( a = 9 \).

### Conclusion:
The value of \( a \) is \( \boxed{9} \).

Would you like further details on any part of the solution or have any questions?

Here are 5 related questions:
1. How do you factor quadratic equations in general?
2. What are the steps for solving equations involving fractions?
3. How does the difference of squares factorization work?
4. Can you explain how to combine rational expressions with different denominators?
5. What are some common mistakes to avoid when solving quadratic equations?

**Tip:** When solving equations with rational expressions, always look for opportunities to factor and simplify terms to reduce complexity.

4/(a^2-a-2) - 2/(a^2+3a+2) = (-3)/(4-a^2)

Learn how to solve the rational expression 4/(a^2 - a - 2) - 2/(a^2 + 3a + 2) = (-3)/(4 - a^2) with a detailed step-by-step solution. This problem covers key algebraic concepts, including factoring quadratics and the difference of squares. Suitable for students in Grades 9-12.

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