Given $I(s)=\dfrac{2s}{(s+1)(s+2)}$

To get poles,equate denominator of I(s) to zero

(s+1)(s+2)=0

poles are s1= -1 and s2 = -2

To get zero, equate numerator of I(s) to zero.

2s=0

zero is s=0

Pole - zero plot

To obtain i(t), need to apply inverse laplase to I(s)

Now Apply Inverse Laplace to I(s)

$L^{-1}(I(s))=L^{-1}(\dfrac{2s}{(s+1)(s+2)})$

we have a inverse laplace formula i.e. $L^{-1}(\dfrac{1}{s+a})=e^{-at}$

now apply partial fractions to $\dfrac{2s}{(s+1)(s+2)}$

Partial Factions:

$\dfrac{2s}{(s+1)(s+2)} = \dfrac{A}{S+1}+\dfrac{B}{s+2}$

$2s=A(s+2)+B(s+1)....................(1)$

take s= -1 in equation (1) to get value of A

2(-1)=A(-1+2)+B(-1+1)

A=-2

take s= -2 in equation (1) to get value of B

2(-2)=A(-2+2)+B(-2+1)

B=4

$\dfrac{2s}{(s+1)(s+2)} = \dfrac{-2}{S+1}+\dfrac{4}{s+2}$

Now apply inverse laplace

$L^{-1}(I(s))=L^{-1}(\dfrac{2s}{(s+1)(s+2)})=L^{-1}\dfrac{-2}{S+1}+L^{-1}\dfrac{4}{s+2}$

by using this formula $L^{-1}(\dfrac{1}{s+a})=e^{-at}$ we get

$i(t)=-2e^{-t}+4e^{-2t}$