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To find the inverse Laplace transform of \( F(s) = \frac{-3s^3 + 6s^2 - 12s + 96}{(s^2 + 4)(s^2 + 16)} \), we should first break down \( F(s) \) into simpler fractions using partial fraction decomposition. Then, we can use standard inverse Laplace transform formulas.

Let's proceed with the steps:

1. **Partial Fraction Decomposition**: Express \( F(s) \) as a sum of simpler fractions. Since the denominator has quadratic terms, we expect terms of the form \( \frac{As + B}{s^2 + 4} \) and \( \frac{Cs + D}{s^2 + 16} \).

2. **Determine Constants**: Solve for \( A \), \( B \), \( C \), and \( D \) by equating coefficients after multiplying out the terms.

3. **Inverse Transform Each Term**: Once decomposed, apply the inverse Laplace transform to each term individually. Standard Laplace transforms like \( \frac{s}{s^2 + a^2} \) and \( \frac{1}{s^2 + a^2} \) can be used.

Would you like me to go through the detailed calculations for each step?

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Here are some related questions to consider:

1. How does partial fraction decomposition help in finding inverse Laplace transforms?
2. What are the standard Laplace transforms of \( \frac{s}{s^2 + a^2} \) and \( \frac{1}{s^2 + a^2} \)?
3. Why do we separate terms involving \( s^2 + 4 \) and \( s^2 + 16 \) in the denominator?
4. Can we use any other methods to simplify this expression before taking the inverse Laplace transform?
5. How can the final answer be verified for accuracy?

**Tip**: Always check the degree of the numerator and denominator to decide the form of partial fraction decomposition.

Find the inverse Laplace transform \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \) of the function \( F(s) = \frac{-3s^3 + 6s^2 - 12s + 96}{(s^2 + 4)(s^2 + 16)} \).

Learn how to find the inverse Laplace transform of \( F(s) = \frac{-3s^3 + 6s^2 - 12s + 96}{(s^2 + 4)(s^2 + 16)} \) using partial fraction decomposition. This solution covers key concepts in Laplace transforms and partial fraction decomposition, suitable for university-level students.

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