Laplace transform of $(t-2)^2u_2$ The homework problem is
$$f(t) = \begin{cases} 0 & t < 2\\ (t-2)^2 & t\geq 2\end{cases}$$
$f(t)$ as a step function
$$f(t) = (t-2)^2u_2(t)$$
Using what we learned in class
$$\mathcal{L}\{f(t)u_c(t)\} = e^{-cs}\mathcal{L}\{f(t+c)\}$$
$\mathcal{L}\{f(t)\}$ should be
$$e^{-2s} \mathcal{L}\{f(t+c)\}= e^{-2s} \frac2{s^3}$$
but the answer in the back of the book is
$$e^{-s} \frac2{s^3}$$
I can't figure out why the answer does not include the 2 in the exponent.

Just, work out the problem from scratch
$$ \int_{0}^{\infty}f(t)e^{-st}dt =  \int_{0}^{\infty}(t-2)^2u_2(t)\,e^{-st} dt = \int_{2}^{\infty}(t-2)^2e^{-st} dt$$ 
$$ = \int_{0}^{\infty}u^2e^{-s(u+2)} du= e^{-2s}\frac{2!}{s^3}, $$
which matches your answer.
