# Inverse Laplace Transform with squared irreducible quadratic in denominator

Okay, the equation is $$\frac{2s^3-2s}{(4s^2-4s+5)^2}$$

So I use partial fractions with $$\frac{As+B}{4s^2-4s+5} + \frac{Cs+D}{(4s^2-4s+5)^2} = 2s^3-2s$$

and square that quadratic get $$(As+B)(16s^4-32s^3+56s^2-40s+25)+(Cs+D)(4s^2-4s+5) = 2s^3-2s$$

This is the part where I'm suppose to plug in values for s to solve for the variables. But the only one that would make part of it $=0$ would make all terms $=0$.

So I tried differentiating both sides some times, each time picking a easy value for s that would equate one variable to $0$. Then I end up with a system of equations to solve, but get the wrong answer. I don't know what I'm doing wrong. Would someone point out what I did wrong or try solving it for me and writing the system of equations they get?

Thanks!

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That should be $\frac{2s^3-2 s}{(4 s^2 - 4 s + 5)^2} = \frac{A s + B}{4 s^2 - 4 s + 5} + \frac{Cs + D}{(4 s^2 - 4 s + 5)^2}$.
Multiplying by the denominator, you should get $2 s^3 - 2 s = (As + B) (4 s^2 - 4 s + 5) + C s + D$.
The "easy" values to substitute are the roots of $4 s^2 - 4 s + 5$, namely $1/2 \pm i$. Or you could use long division with remainder.