The given image shows a partial fraction decomposition problem, and it appears to be:

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

To solve this, we need to find the constants $A$, $B$, and $C$.

Step-by-step Solution:

1. Multiply both sides by the denominator $(s + 4)^2(s - 1)$ to eliminate the fractions:

$s - 4 = A(s + 4)(s - 1) + B(s - 1) + C(s + 4)^2$

2. Expand both sides:

   • Expand $A(s + 4)(s - 1)$: $A(s + 4)(s - 1) = A(s^2 - s + 4s - 4) = A(s^2 + 3s - 4)$
   • Expand $B(s - 1)$: $B(s - 1) = B(s - 1)$
   • Expand $C(s + 4)^2$: $C(s + 4)^2 = C(s^2 + 8s + 16)$

   Now, the equation becomes: $s - 4 = A(s^2 + 3s - 4) + B(s - 1) + C(s^2 + 8s + 16)$

3. Group similar terms: $s - 4 = (A + C)s^2 + (3A + B + 8C)s + (-4A - B + 16C)$

4. Compare coefficients: Now, we compare the coefficients of powers of $s$ on both sides of the equation:

   • For $s^2$: $A + C = 0$
   • For $s$: $3A + B + 8C = 1$
   • Constant term: $-4A - B + 16C = -4$

5. Solve the system of equations:

   From $A + C = 0$, we get: $A = -C$

   Substitute $A = -C$ into the other two equations:

   For $3A + B + 8C = 1$: $3(-C) + B + 8C = 1 \quad \Rightarrow \quad -3C + B + 8C = 1 \quad \Rightarrow \quad B + 5C = 1 \quad \Rightarrow \quad B = 1 - 5C$

   For $-4A - B + 16C = -4$: $-4(-C) - B + 16C = -4 \quad \Rightarrow \quad 4C - B + 16C = -4 \quad \Rightarrow \quad 20C - B = -4$

   Substitute $B = 1 - 5C$ into this equation: $20C - (1 - 5C) = -4 \quad \Rightarrow \quad 20C - 1 + 5C = -4 \quad \Rightarrow \quad 25C = -3 \quad \Rightarrow \quad C = -\frac{3}{25}$

6. Find $A$ and $B$:

   • Since $A = -C$, we have: $A = \frac{3}{25}$
   • Using $B = 1 - 5C$: $B = 1 - 5\left(-\frac{3}{25}\right) = 1 + \frac{15}{25} = 1 + \frac{3}{5} = \frac{8}{5}$

Final Answer:

Thus, the partial fraction decomposition is: $\frac{s - 4}{(s + 4)^2 (s - 1)} = \frac{\frac{3}{25}}{s + 4} + \frac{\frac{8}{5}}{(s + 4)^2} + \frac{-\frac{3}{25}}{s - 1}$

Let me know if you'd like further details or have any questions!

Here are 5 related questions you might explore:

1. How can you solve partial fractions with more than two terms in the denominator?
2. What happens if the numerator is a quadratic term instead of linear?
3. How do partial fractions help in solving integrals?
4. Can partial fractions be applied to improper fractions, and how do you handle them?
5. What is the significance of the degree of the denominator in determining the form of the partial fraction?

Tip: When solving partial fractions, always expand and match coefficients of each power of the variable to set up a system of equations efficiently.