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The image you have shared contains a partial fraction decomposition problem and the related inverse Laplace transform.

The problem starts with the function \( F(s) \), given by:

\[
F(s) = \frac{-5s + 5}{s^3 + 8s^2 + 20s + 16}
\]

The task is to decompose it into partial fractions and then find the corresponding inverse Laplace transform.

Let’s proceed step by step:

### Step 1: Factoring the denominator
First, the denominator of \( F(s) \) is:

\[
s^3 + 8s^2 + 20s + 16
\]

We need to factor this cubic polynomial. Based on common factorization techniques, this expression can be factored as:

\[
(s + 4)(s + 2)^2
\]

### Step 2: Partial Fraction Decomposition
Now, we express \( F(s) \) as a sum of simpler fractions:

\[
F(s) = \frac{-5s + 5}{(s + 4)(s + 2)^2} = \frac{A}{s + 4} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2}
\]

We need to determine the values of \( A \), \( B \), and \( C \) by multiplying through by the denominator and solving for the constants.

### Step 3: Inverse Laplace Transform
Once we have the partial fraction decomposition, we can find the inverse Laplace transform of each term separately.

The time-domain function \( f(t) \) is then:

\[
f(t) = \mathcal{L}^{-1} \left( \frac{A}{s + 4} \right) + \mathcal{L}^{-1} \left( \frac{B}{s + 2} \right) + \mathcal{L}^{-1} \left( \frac{C}{(s + 2)^2} \right)
\]

Using standard inverse Laplace transforms:

- \( \mathcal{L}^{-1} \left( \frac{A}{s + a} \right) = A e^{-at} \)
- \( \mathcal{L}^{-1} \left( \frac{B}{(s + a)^2} \right) = B t e^{-at} \)

The inverse transform of the entire function will combine these terms.

Let me calculate the exact values of \( A \), \( B \), and \( C \) for you.The partial fraction decomposition of \( F(s) \) is:

\[
F(s) = \frac{25}{4(s + 4)} - \frac{25}{4(s + 2)} + \frac{15}{2(s + 2)^2}
\]

### Step 4: Inverse Laplace Transform
Using the inverse Laplace transforms of these terms:

1. \( \mathcal{L}^{-1} \left( \frac{25}{4(s + 4)} \right) = \frac{25}{4} e^{-4t} \)
2. \( \mathcal{L}^{-1} \left( \frac{25}{4(s + 2)} \right) = \frac{25}{4} e^{-2t} \)
3. \( \mathcal{L}^{-1} \left( \frac{15}{2(s + 2)^2} \right) = \frac{15}{2} t e^{-2t} \)

Thus, the time function \( f(t) \) is:

\[
f(t) = \frac{25}{4} e^{-4t} - \frac{25}{4} e^{-2t} + \frac{15}{2} t e^{-2t}
\]

Would you like any further details on this solution or how to apply these concepts? 

### Related Questions:
1. How do you factor higher-degree polynomials in general?
2. What is the method to solve partial fraction decompositions for repeated roots?
3. How do you compute the inverse Laplace transform of more complex functions?
4. What are the key applications of the Laplace transform in control theory?
5. How does the Laplace transform relate to solving differential equations?

### Tip:
When dealing with repeated roots in partial fraction decomposition, always include terms with increasing powers of the root in the denominator (e.g., \( \frac{B}{(s+a)^2} \)).

The partial-fraction expansion of F(s) = (-5s + 5) / (s^3 + 8s^2 + 20s + 16) is sought, along with the corresponding inverse Laplace transform f(t).

Solve for the partial fraction expansion of F(s) = (-5s + 5) / (s^3 + 8s^2 + 20s + 16) and its inverse Laplace transform f(t). This problem covers advanced concepts in control theory and Laplace transforms, suitable for undergraduate students.

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