To express $\frac{x + 3}{(x - 2)(x + 4)}$ as partial fractions, we proceed as follows:

1. Set up the partial fraction decomposition: Since the denominator has two distinct linear factors, $x - 2$ and $x + 4$, we can write: $\frac{x + 3}{(x - 2)(x + 4)} = \frac{A}{x - 2} + \frac{B}{x + 4}$ where $A$ and $B$ are constants that we need to determine.

2. Multiply through by the denominator $(x - 2)(x + 4)$ to clear the fractions: $x + 3 = A(x + 4) + B(x - 2)$

3. Expand and collect terms: Expanding both sides, we get: $x + 3 = Ax + 4A + Bx - 2B$ Grouping the terms by powers of $x$, this becomes: $x + 3 = (A + B)x + (4A - 2B)$

4. Set up equations by equating coefficients: For the coefficients of $x$ and the constant terms, we have:

   • For $x$: $A + B = 1$
   • For the constant term: $4A - 2B = 3$

5. Solve the system of equations: From $A + B = 1$, we get $B = 1 - A$.

   Substitute $B = 1 - A$ into the second equation: $4A - 2(1 - A) = 3$ Simplify: $4A - 2 + 2A = 3$ $6A = 5$ $A = \frac{5}{6}$

   Substitute $A = \frac{5}{6}$ into $B = 1 - A$: $B = 1 - \frac{5}{6} = \frac{1}{6}$

6. Write the final partial fraction decomposition: $\frac{x + 3}{(x - 2)(x + 4)} = \frac{5}{6(x - 2)} + \frac{1}{6(x + 4)}$

Would you like further details on any step or have any questions?

Follow-up Questions

1. What are partial fractions used for in calculus?
2. How would you handle partial fraction decomposition if the denominator had a repeated factor?
3. Can partial fraction decomposition be applied to a denominator with irreducible quadratic terms?
4. How would you solve for $A$ and $B$ using a different method, like substituting specific values for $x$?
5. How does partial fraction decomposition help in integrating rational functions?

Tip: Always check that the degree of the numerator is lower than the degree of the denominator before attempting partial fraction decomposition.