Laplace transform (Using tables) of following function, is this correct?

Question

$$ 4H(t-3)~. t^2$$

For my answer I used the the t-shift theorem (using tables) and I got
$$ 4e^{-3s} . \dfrac{2}{s^3} $$

I have a feeling this isn't correct.

Answer 1

We need to write $t^2$ in terms of $t-3$ to use the Shifting Property or Laplace Transform definition (integral instead of tables).

We have

$$t^2 = ((t-3) + 3)^ 2 = (t-3)^2 + 6 (t-3) + 9$$

Hence, we can write

$$4 H(t-3)t^2 = 4 H(t-3)((t-3)^2 + 6 (t-3) + 9)$$

Using those three pieces, we end up with

$$\mathcal{L}\left(\ 4~H(t-3)~ t^2\right) = \dfrac{4 e^{-3 s} \left(9 s^2+6 s+2\right)}{s^3}$$