Inverse laplace transform using convolution theorem

can some one please find L^-1[16/((s-2)(s+4))] by convolution theorem

The Laplace transform of an exponential function known as:
ℒ{ e^(-a∙t) } = 1/(s + a)

So your transform can be written as product of the transform of two exponential functions:
16/((s - 2)∙(s + 4)) = 16∙ℒ{ e^(2∙t) }∙ℒ{ e^(-4∙t) }

Convolution theorem states that
ℒ{ ∫[0;t] f(u)∙g(t-u) du } = ℒ{ f(t) }∙ℒ{ g(t) }

Hence,
ℒ⁻¹{ 16/((s - 2)∙(s + 4)) }
= 16 ∙ ℒ⁻¹{ ℒ{ e^(2∙t) }∙ℒ{ e^(-4∙t) } }
= 16 ∙ ∫[0;t] e^(2∙u) ∙ e^(-4∙t)∙ e^(4∙u) ) du
= 16∙e^(-4∙t) ∙ ∫[0;t] e^(2∙u)∙e^(4∙u) ) du
= 16∙e^(-4∙t) ∙ ∫[0;t] e^(6∙u) du
= 16∙e^(-4∙t) ∙ ( (1/6)∙e^(6∙t) - (1/6)∙e^(6∙0) )
= (8/3)∙e^(-4∙t) ∙ (e^(6∙t) - 1)
= (8/3)∙e^(4∙t) - (8/3)∙e^(-4∙t)