Factor the denominator by factor theorem
$f(s) = s^3 - 3s^2 - 13s + 15$
$f(1) = 1^3 - 3(1^2) - 13(1) + 15 = 0$ → $(s - 1)$ is a factor
$f(-1) = (-1)^3 - 3(-1)^2 - 13(-1) + 15 = 24$
$f(3) = 3^3 - 3(3^2) - 13(3) + 15 = -24$
$f(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15 = 0$ → $(s + 3)$ is a factor
$f(5) = 5^3 - 3(5^2) - 13(5) + 15 = 0$ → $(s - 5)$ is a factor
$s^3 - 3s^2 - 13s + 15 = (s - 1)(s + 3)(s - 5)$
Thus,
$\dfrac{2s^2 + 5s - 6}{s^3 - 3s^2 - 13s + 15} = \dfrac{2s^2 + 5s - 6}{(s - 1)(s + 3)(s - 5)}$
For
$\dfrac{2s^2 + 5s - 6}{(s - 1)(s + 3)(s - 5)} = \dfrac{A}{s - 1} + \dfrac{B}{s + 3} + \dfrac{C}{s - 5}$
$2s^2 + 5s - 6 = A(s + 3)(s - 5) + B(s - 1)(s - 5) + C(s - 1)(s + 3)$
Set $s = 1$
$1 = -16A$
$A = -1/16$
Set $s = -3$
$-3 = 32B$
$B = -3/32$
Set $s = 5$
$69 = 32C$
$C = 69/32$
Therefore,
$\mathcal{L}^{-1} \left[ \dfrac{2s^2 + 5s - 6}{s^3 - 3s^2 - 13s + 15} \right] = \mathcal{L}^{-1} \left[ \dfrac{-1/16}{s - 1} + \dfrac{-3/32}{s + 3} + \dfrac{69/32}{s - 5} \right]$
$\mathcal{L}^{-1} \left[ \dfrac{2s^2 + 5s - 6}{s^3 - 3s^2 - 13s + 15} \right] = -\dfrac{1}{16}\mathcal{L}^{-1} \left( \dfrac{1}{s - 1} \right) - \dfrac{3}{32} \mathcal{L}^{-1} \left( \dfrac{1}{s + 3} \right) + \dfrac{69}{32} \mathcal{L}^{-1} \left( \dfrac{1}{s - 5} \right)$
$\mathcal{L}^{-1} \left[ \dfrac{2s^2 + 5s - 6}{s^3 - 3s^2 - 13s + 15} \right] = -\frac{1}{16}e^t - \frac{3}{32} e^{-3t} + \frac{69}{32} e^{5t}$ answer
