To find the partial fractions of the expression $\frac{3x + 5}{(x - 3)(2x + 1)}$, we can break it down into separate terms in the form:

$\frac{3x + 5}{(x - 3)(2x + 1)} = \frac{A}{x - 3} + \frac{B}{2x + 1}$

where $A$ and $B$ are constants that we need to determine.

Step 1: Multiply through by the denominator

Multiply both sides by $(x - 3)(2x + 1)$ to clear the fraction:

$3x + 5 = A(2x + 1) + B(x - 3)$

Step 2: Expand and Simplify

Expanding the right side gives:

$3x + 5 = 2Ax + A + Bx - 3B$

Combine like terms:

$3x + 5 = (2A + B)x + (A - 3B)$

Step 3: Set up equations by comparing coefficients

By comparing coefficients of $x$ and the constant terms, we get two equations:

1. For the $x$ terms: $2A + B = 3$
2. For the constant terms: $A - 3B = 5$

Step 4: Solve the system of equations

Using substitution or elimination, we can solve for $A$ and $B$:

1. From $2A + B = 3$,
2. $A - 3B = 5$.

After solving, we find that $A = 1$ and $B = 5$.

Final Answer

The partial fraction decomposition is:

$\frac{3x + 5}{(x - 3)(2x + 1)} = \frac{1}{x - 3} + \frac{5}{2x + 1}$

Thus, the correct answer is: Option (c): $\frac{1}{x - 3} + \frac{5}{2x + 1}$.

Would you like more details on the solution steps or have any other questions?

Related Questions

1. How do you find partial fraction decompositions for higher-degree polynomials?
2. What are the applications of partial fractions in calculus, especially in integration?
3. How do you handle repeated roots in partial fraction decompositions?
4. Can you explain the method of partial fractions in cases with irreducible quadratic factors?
5. What happens if the degree of the numerator is higher than the denominator in partial fraction decomposition?

Tip

When comparing coefficients in partial fractions, ensure that each term's power of $x$ matches between both sides to set up accurate equations.