I am trying to find the inverse Laplace transformation of the following equation, specifically factorising the denominator

Question

The function has poles at s = ±1, \[ F (s) = {8 \over s^4 - 2s^3 + 2s - 1} \]

The issue I'm having is that I cannot find a way to factorise the denominator so that I can use the Heaviside method. I would massively appreciate if anyone could point me in the direction of how to do it.

Answer 1

Letting $P(X):=X^4-2X^3+2X-1$. Notice that the possible rational roots are $X=\pm 1$.

Now,

• $P(1)=0\implies X=1$ root.
• $P(-1)=0\implies X=-1$ root.

Hence $(X-1)$ and $(X+1)$ are factor of $P(X)$ by factor theorem.

• Try long division give with $(X-1)$

$$\frac{P(X)}{(X-1)}=X^3-X^2-X+1:=Q(X)\implies P(X)=(X-1)Q(X)$$

• Try long division with $(X+1)$ $$\frac{Q(X)}{(X+1)}=X^2-2X+1:=R(X)\implies Q(X)=(X+1)R(X)$$

But $$X^2-2X+1=(X-1)(X-1)\iff R(X)=(X-1)^2$$ We are lucky, it all factored into $\mathbf{R}[X]$.

Hence $$X^4-2X^3+2X-1=(X-1)^{3}(X+1)$$ Therefore $$\frac{1}{8}F(S)=\frac{1}{S^4-2S^3+2S-1}=\frac{1}{(S-1)^3(S+1)}$$

One can recognize directly from the table of Laplace transforms the answer

$$f(t)=e^{t}(2t^2-2t+1)-e^{-t}.$$