clarification with inverse Laplace transform

Question

$L[(A)]=\frac{1}{p(p^2+p-6)}$

$A=-\frac{1}{6}+\frac{1}{15}e^{-3t}+\frac{1}{10}e^{2t}$

Could someone please explain how the coefficients, $-\frac{1}{6}, \frac{1}{15}, \frac{1}{10}$ were obtained? I am unable to understand this.

Answer 1

It is simply a PFE $$f(p)=\frac{1}{p(p^2+p-6)}=\frac{1}{p(p-2)(p+3)}=\frac{A}{p}+\frac{B}{p-2}+\frac{C}{p+3}$$ $$A=pf(0)=\frac{1}{(-2)(3)}=-\frac{1}{6}$$ $$B=(p-2)f(2)=\frac{1}{(2)(2+3)}=\frac{1}{10}$$ $$C=(p+3)f(-3)=\frac{1}{(-3)(-3-2)}=\frac{1}{15}$$