We are given the function $p(x) = \frac{2x}{(x + 2)^2}, x \neq -2$, and we are tasked to express it as a sum of two terms of the form $\frac{A}{x + 2} + \frac{B}{(x + 2)^2}$, and then find the constants $A$ and $B$.

Step 1: Set up the equation

We want to match the given expression with:

$p(x) = \frac{A}{x + 2} + \frac{B}{(x + 2)^2}.$

Start by rewriting the right-hand side with a common denominator:

$\frac{A}{x + 2} + \frac{B}{(x + 2)^2} = \frac{A(x + 2) + B}{(x + 2)^2}.$

Now, equate this to the given expression for $p(x)$:

$\frac{A(x + 2) + B}{(x + 2)^2} = \frac{2x}{(x + 2)^2}.$

Step 2: Compare numerators

Since the denominators are the same, we can equate the numerators:

$A(x + 2) + B = 2x.$

Step 3: Expand and group terms

Now expand the left-hand side:

$A(x + 2) + B = Ax + 2A + B.$

This gives us:

$Ax + 2A + B = 2x.$

Step 4: Compare coefficients

Now, compare the coefficients of $x$ and the constant terms on both sides of the equation:

• Coefficient of $x$: $A = 2$,
• Constant term: $2A + B = 0$.

Step 5: Solve for $A$ and $B$

We already have $A = 2$. Substitute $A = 2$ into the second equation:

$2(2) + B = 0 \implies 4 + B = 0 \implies B = -4.$

Final Answer:

The values of $A$ and $B$ are:

$A = 2, \quad B = -4.$

Would you like more details, or do you have any other questions?

Related Questions:

1. How can we check if the values of $A$ and $B$ are correct?
2. What method can we use for partial fraction decomposition for higher-order polynomials?
3. Can this technique be used for improper rational functions?
4. How would you handle a situation where the denominator is not factored easily?
5. How does this process change if the power of $(x + 2)$ in the denominator was higher than 2?

Tip: Always expand and simplify both sides of the equation when working with partial fractions to ensure the terms align correctly.