To resolve the given rational expression into partial fractions, we start by analyzing the expression:

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

Since the denominator consists of two linear factors, $(x+4)$ and $(x+5)$, we can express the fraction as the sum of two partial fractions:

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

Step 1: Set up the equation

We multiply both sides of the equation by $(x+4)(x+5)$ to eliminate the denominators:

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

Step 2: Expand the right-hand side

Now expand both terms on the right-hand side:

$A(x+5) = A x + 5A$ $B(x+4) = B x + 4B$

Thus, the equation becomes:

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

Step 3: Compare coefficients

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

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

Step 4: Solve the system of equations

We now solve this system of equations:

1. $A + B = 3$
2. $5A + 4B = 5$

Solve the first equation for $A$:

$A = 3 - B$

Substitute this into the second equation:

$5(3 - B) + 4B = 5$ $15 - 5B + 4B = 5$ $15 - B = 5$ $B = 10$

Now substitute $B = 10$ back into $A + B = 3$:

$A + 10 = 3$ $A = -7$

Step 5: Write the partial fractions

Thus, the partial fraction decomposition is:

$\frac{2x^2 + 3x + 5}{(x+4)(x+5)} = \frac{-7}{x+4} + \frac{10}{x+5}$

Would you like any further clarification or a more detailed explanation?

5 Related Questions:

1. What are partial fractions used for in integration?
2. How do you decompose fractions with repeated linear factors?
3. What happens if the numerator has a higher degree than the denominator?
4. How would the method change for irreducible quadratic factors?
5. Can partial fractions be applied to complex numbers?

Tip:

Always check if the numerator's degree is less than the denominator before applying partial fractions. If not, perform polynomial long division first.