Answer:

Explanation:
![L^(-1)[(2s+4)/((s-3)^(3)) ]=](https://img.qammunity.org/2020/formulas/mathematics/college/rzgoym4k4md6zuctztolrfu8jn5fteb9xe.png)
Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3
Translation theorem:
![L^(-1)[(2s+4)/((s-3)^(3)) ]=e^(3t) L^(-1)[(2(s+3)+4)/(s^(3)) ]](https://img.qammunity.org/2020/formulas/mathematics/college/sh1jh9ru0j641gju4ba84sznfz99a0quzy.png)
Separate the fraction in a sum:
![e^(3t) L^(-1)[(2s+10)/(s^(3)) ]=e^(3t) L^(-1)[(2s)/(s^(3))+(10)/(s^(3)) ]=e^(3t) (L^(-1)[(2)/(s^(2))]+ L^(-1)[(10)/(s^(3))])](https://img.qammunity.org/2020/formulas/mathematics/college/9q95hv42wuv9mqxv7xv5uljpitzypc3i5w.png)
The formula for this is:
![L^(-1)[(n!)/(s^(n+1)) ]=t^(n)](https://img.qammunity.org/2020/formulas/mathematics/college/m7br4b22eb0nzyiary7m75ehmbwwtnig6x.png)
Modify the expression to match the formula.
![e^(3t) (2L^(-1)[(1)/(s^(1+1))]+ (10)/(2) L^(-1)[(2)/(s^(2+1))])=e^(3t) (2L^(-1)[(1)/(s^(1+1))]+ 5 L^(-1)[(2)/(s^(2+1))])](https://img.qammunity.org/2020/formulas/mathematics/college/6r473flpwv14fo69ydyudnvknxgh8bm7kd.png)
Solve
![e^(3t) (2L^(-1)[(1)/(s^(1+1))]+ 5 L^(-1)[(2)/(s^(2+1))])=e^(3t)(2t+5t^(2) )](https://img.qammunity.org/2020/formulas/mathematics/college/tk1na96d9fuyq6sofkdr6pehqnwr45na7h.png)