kindly solve using the second shifting theorem. thanks. please include explanation that I can understand. many thanks. L[t^2 u(t – 3)] -3)]

Answer:[tex]\dfrac{2}{s^3}e^{-3s}\ +\ \dfrac{6}{s^2}e^{-3s}\ +\ \dfrac{9}{s}e^{-3s}\ -\ \dfrac{3}{s}[/tex]Step-by-step explanation:Given polynomial,[tex]f(t)\ =\ t^2.u(t-3)\ -\ 3[/tex]we can write above polynomial as[tex]f(t)\ =\ (t-3+3)^2.u(t-3)\ -\ 3[/tex]      [tex]=\ ((t-3)^2\ +\ 2\times 3\times (t-3)\ +\ 3^2).u(t-3)-3[/tex]      [tex]=\ (t-3)^2.u(t-3)\ +\ 6(t-3).u(t-3)\ +\ 9.u(t-3)\ -\ 3[/tex]Now, we have to calculate the Laplace of above polynomialaccording to shifting property of Laplace transform, we can write[tex]f(t-t_0)\ =\ F(s).e^{-st_0}[/tex]So, we can write the Laplace transform of above polynomial as[tex]L[f(t)]\ =\ L[(t-3)^2.u(t-3)\ +\ 6(t-3).u(t-3)\ +\ 9.u(t-3)\ -\ 3][/tex]           [tex]=\ \dfrac{2}{s^3}e^{-3s}\ +\ \dfrac{6}{s^2}e^{-3s}\ +\ \dfrac{9}{s}e^{-3s}\ -\ \dfrac{3}{s}[/tex]So, the Laplace transform of the given polynomial will be[tex]\ \dfrac{2}{s^3}e^{-3s}\ +\ \dfrac{6}{s^2}e^{-3s}\ +\ \dfrac{9}{s}e^{-3s}\ -\ \dfrac{3}{s}[/tex]