To decompose the given expression into partial fractions, we start with the following rational function:

$\frac{16 + 5x - 2x^2}{(x+1)^2 (x+4)}$

Step 1: Set up the form for partial fractions

Since the denominator has the factors $(x+1)^2$ and $(x+4)$, the corresponding partial fraction decomposition will be:

$\frac{16 + 5x - 2x^2}{(x+1)^2 (x+4)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+4}$

where $A$, $B$, and $C$ are constants to be determined.

Step 2: Combine the fractions on the right-hand side

To combine these fractions over a common denominator $(x+1)^2 (x+4)$, we have:

$\frac{A}{x+1} = \frac{A(x+1)(x+4)}{(x+1)^2(x+4)}$ $\frac{B}{(x+1)^2} = \frac{B(x+4)}{(x+1)^2(x+4)}$ $\frac{C}{x+4} = \frac{C(x+1)^2}{(x+1)^2(x+4)}$

Now, summing these terms, we get:

$\frac{A(x+4) + B(x+4) + C(x+1)^2}{(x+1)^2(x+4)}$

Step 3: Set up the equation

The numerator of the original fraction $16 + 5x - 2x^2$ must equal the numerator of the combined partial fractions:

$16 + 5x - 2x^2 = A(x+4) + B(x+4) + C(x+1)^2$

Expand both sides to find $A$, $B$, and $C$.

Left-hand side:

$16 + 5x - 2x^2$

Right-hand side:

Expanding the right-hand side: $A(x+4) + B(x+4) + C(x+1)^2 = A(x+4) + B(x+4) + C(x^2 + 2x + 1)$ $= (A+B)x + (4A + 4B) + C(x^2 + 2x + 1)$ $= Cx^2 + (A + B + 2C)x + (4A + 4B + C)$

Step 4: Equate coefficients

Now, compare coefficients of powers of $x$ on both sides:

• Coefficient of $x^2$: $-2 = C$, so $C = -2$
• Coefficient of $x$: $5 = A + B + 2C = A + B + 2(-2) = A + B - 4$, so $A + B = 9$
• Constant term: $16 = 4A + 4B + C = 4A + 4B - 2$, so $4A + 4B = 18$, or $A + B = \frac{9}{2}$

Step 5: Solve the system of equations

We now solve for $A$ and (B: