The expression you've provided is:

$\frac{x^2 + 3x + 4}{(2x + 2)(2x^2 + 5)}$

To resolve this expression into partial fractions, we will express it in the form:

$\frac{x^2 + 3x + 4}{(2x + 2)(2x^2 + 5)} = \frac{A}{2x + 2} + \frac{Bx + C}{2x^2 + 5}$

Here, $A$, $B$, and $C$ are constants that we need to determine.

Step 1: Clear the denominator

Multiply both sides by the denominator $(2x + 2)(2x^2 + 5)$:

$x^2 + 3x + 4 = A(2x^2 + 5) + (Bx + C)(2x + 2)$

Step 2: Expand and collect like terms

Expand the right side:

$x^2 + 3x + 4 = A(2x^2 + 5) + (Bx + C)(2x + 2)$

Expanding further:

$x^2 + 3x + 4 = A \cdot 2x^2 + 5A + Bx \cdot 2x + Bx \cdot 2 + C \cdot 2x + C \cdot 2$

$x^2 + 3x + 4 = 2Ax^2 + 5A + 2Bx^2 + 2Bx + 2Cx + 2C$

Combine like terms:

$x^2 + 3x + 4 = (2A + 2B)x^2 + (2B + 2C)x + (5A + 2C)$

Step 3: Equate coefficients

Now, equate the coefficients of corresponding powers of $x$ on both sides:

• Coefficient of $x^2$: $2A + 2B = 1$
• Coefficient of $x$: $2B + 2C = 3$
• Constant term: $5A + 2C = 4$

Step 4: Solve the system of equations

Let's solve these equations:

1. $2A + 2B = 1 \Rightarrow A + B = \frac{1}{2}$
2. $2B + 2C = 3 \Rightarrow B + C = \frac{3}{2}$
3. $5A + 2C = 4$

From the first equation:

$B = \frac{1}{2} - A$

Substitute $B$ in the second equation:

$\frac{1}{2} - A + C = \frac{3}{2} \Rightarrow C = A + 1$

Now, substitute $C$ into the third equation:

$5A + 2(A + 1) = 4 \Rightarrow 7A + 2 = 4 \Rightarrow 7A = 2 \Rightarrow A = \frac{2}{7}$

Substitute $A$ into the expressions for $B$ and $C$:

$B = \frac{1}{2} - \frac{2}{7} = \frac{7}{14} - \frac{4}{14} = \frac{3}{14}$

$C = \frac{2}{7} + 1 = \frac{2}{7} + \frac{7}{7} = \frac{9}{7}$

Final Answer

So, the partial fraction decomposition of the given expression is:

$\frac{x^2 + 3x + 4}{(2x + 2)(2x^2 + 5)} = \frac{\frac{2}{7}}{2x + 2} + \frac{\frac{3}{14}x + \frac{9}{7}}{2x^2 + 5}$

Would you like more details or have any questions?

---

Here are five related questions to consider:

1. How would the partial fraction decomposition change if the numerator was $x^2 + 3x + 1$ instead of $x^2 + 3x + 4$?
2. Can you solve the partial fraction decomposition if the denominator were $(2x + 1)(x^2 + 2)$?
3. What are the implications if the quadratic in the numerator is factorable?
4. How do the constants $A$, $B$, and $C$ affect the shape of the graph of the partial fractions?
5. How can you confirm that the partial fraction decomposition is correct?

Tip: Always double-check your expanded expressions and coefficients before solving the system of equations in partial fractions.