Laplace transform polynomial fraction

Question

Can anybody help, how do I find the inverse Laplace transform of $\frac{1}{s^3 + 3s^2 + 2s + 4}$?

Answer 1

As said in comments, the factorization is not easy, in particular because the equation shows two non(real complex conjugate roots.

Assuming that you know how to work with the reciprocal of a quadratic polynomial, we can write $$s^3 + 3s^2 + 2s + 4=(s+1+k) \Big(s^2+(2-k) s+\frac 4{1+k}\Big)$$ where $$k=\frac{2 }{\sqrt{3}}\cosh \left(\frac{1}{3} \cosh ^{-1}\left(6 \sqrt{3}\right)\right)$$ and the partial fraction decomposition woudl give $$\frac{1}{s^3 + 3s^2 + 2s + 4}=\frac{1}{2 k^3+3 k^2+3}\Big( \frac{k+1}{s+k+1}+\frac{\left(2 k^3+3 k^2-1\right)-(k+1)^2 s } {4+\left(-k^2+k+2\right) s+(k+1) s^2 } \Big)$$