<s>[INST] Find the Laplace transform of $f(t) = \begin{cases} 0, & \text{if $t<5$} \\ t^2−10t+31, & \text{if $t\ge 5$} \\ \end{cases} $ Find the Laplace transform of
$$f(t) =
\begin{cases}
0,  & \text{if $t<5$} \\
t^2−10t+31, & \text{if $t\ge 5$}  \\
\end{cases}
$$
$F(s)=$ __________?
Here is my work. I went wrong somewhere. Can someone tell me the correct answer
First, rewrite $t^2 - 10t + 31$ in powers of $t - 5$:
$$\begin{split}
  t^2 - 10t + 31
&= [(t - 5) + 5]^2 - 10 [(t - 5) + 5] + 31 \\
&= [(t - 5)^2 + 10(t - 5) + 25] - [10(t - 5) + 50] + 31 \\
&= (t - 5)^2 + 1.
\end{split}
$$
Hence,
$$\begin{split}
f(t) &= ((t - 5)^2 + 1) u(t - 5), \text{which implies}\\
F(s) &= \frac{2!}{s^3} + \frac{\exp(-5s)}{s} 
\end{split}$$
by the shifting theorem.
 [/INST] Your simplification is wrong. Note that 
$(t-5)^2+1 = t^2 -10t + 26$ and you should have $t^2-10t+31$, so you need to use $(t-5)^2+6$ instead.
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