Resolve into Partial Fractions
(vi) `(x^(2)+13x+15)/((2x+3)(x+3)^(2))`

Answer

To resolve the expression \(\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2}\) into partial fractions, we will follow these steps:

### Step 1: Set up the partial fraction decomposition
We can express the given fraction as:
\[ \frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = \frac{A}{2x + 3} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2} \]
where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 2: Clear the denominators
Multiply both sides by the denominator \((2x + 3)(x + 3)^2\):
\[ x^2 + 13x + 15 = A(x + 3)^2 + B(2x + 3)(x + 3) + C(2x + 3) \]

### Step 3: Expand the right-hand side
Now, we will expand the right-hand side:
1. Expand \(A(x + 3)^2\):
\[ A(x^2 + 6x + 9) = Ax^2 + 6Ax + 9A \]
2. Expand \(B(2x + 3)(x + 3)\):
\[ B(2x^2 + 9x + 9) = 2Bx^2 + 9Bx + 9B \]
3. Expand \(C(2x + 3)\):
\[ C(2x + 3) = 2Cx + 3C \]

Combining these, we have:
\[ x^2 + 13x + 15 = (A + 2B)x^2 + (6A + 9B + 2C)x + (9A + 9B + 3C) \]

### Step 4: Set up equations by comparing coefficients
Now, we will compare coefficients from both sides:
1. For \(x^2\):
\[ A + 2B = 1 \quad \text{(1)} \]
2. For \(x\):
\[ 6A + 9B + 2C = 13 \quad \text{(2)} \]
3. For the constant term:
\[ 9A + 9B + 3C = 15 \quad \text{(3)} \]

### Step 5: Solve the system of equations
From equation (1):
\[ A = 1 - 2B \quad \text{(4)} \]

Substituting (4) into (2):
\[ 6(1 - 2B) + 9B + 2C = 13 \]
\[ 6 - 12B + 9B + 2C = 13 \]
\[ -3B + 2C = 7 \quad \text{(5)} \]

Now substitute (4) into (3):
\[ 9(1 - 2B) + 9B + 3C = 15 \]
\[ 9 - 18B + 9B + 3C = 15 \]
\[ -9B + 3C = 6 \quad \text{(6)} \]

Now we have a simpler system of equations (5) and (6):
1. \(-3B + 2C = 7\)
2. \(-9B + 3C = 6\)

From (5), solve for \(C\):
\[ 2C = 7 + 3B \quad \Rightarrow \quad C = \frac{7 + 3B}{2} \quad \text{(7)} \]

Substituting (7) into (6):
\[ -9B + 3\left(\frac{7 + 3B}{2}\right) = 6 \]
\[ -9B + \frac{21 + 9B}{2} = 6 \]
Multiplying through by 2 to eliminate the fraction:
\[ -18B + 21 + 9B = 12 \]
\[ -9B = -9 \quad \Rightarrow \quad B = 1 \]

Now substitute \(B = 1\) back into (4):
\[ A = 1 - 2(1) = -1 \]

Now substitute \(B = 1\) into (7) to find \(C\):
\[ C = \frac{7 + 3(1)}{2} = \frac{10}{2} = 5 \]

### Step 6: Write the final partial fraction decomposition
Now we have:
\[ A = -1, \quad B = 1, \quad C = 5 \]
Thus, the partial fraction decomposition is:
\[ \frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = \frac{-1}{2x + 3} + \frac{1}{x + 3} + \frac{5}{(x + 3)^2} \]