# Having trouble using my usual method of partial fraction decomposition for $\frac{9 + 3s}{s^3 + 2s^2 - s - 2}$.

I'm having trouble using my usual method of partial fraction decomposition for $$\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2}$$.

We can factor such that $$\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2} = \dfrac{A}{s - 1} + \dfrac{B}{s + 1} + \dfrac{C}{s + 2}$$

Therefore, $$9 + 3s = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1)$$

And we have that $$s \not = 1, -1, -2$$.

From here I usually plug in values to find $$A$$, $$B$$, and $$C$$.

$$s = 0:$$

$$9 = 2A - 2B - C \implies C = 2A - 2B - 9$$

$$s = 2:$$

$$15 = 12A + 4B + 3C \implies A = \dfrac{15 - 4B - 3C}{12}$$

$$s = 3:$$

$$18 = 20A + 10B + 8C \implies 10B = 18 - 20A - 8C \implies B = \dfrac{9}{5} - 2A - \dfrac{4C}{5}$$

We now have equations for $$A$$, $$B$$, and $$C$$. But if I try to substitute them into each other, this will result in an infinite loop of substitution.

This method for partial fraction decomposition have always worked for me in the past, so I don't understand why it isn't working in this situation.

I would greatly appreciate it if people could please take the time to explain why my method for partial fraction decomposition is not working in this case and what I should do.

• Try expanding the RHS, collecting like terms and then equating terms on LHS/RHS side of $$3 s + 9 = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1)$$ You get 3-equations and 3 unknowns and arrive at $$A = 2, B = -3, C = 1$$ – Moo May 13 '17 at 11:51
• @Moo I think I found my mistake. For some reason, I've confused myself by convincing myself that $s \not = 1, -1, -2$; but $s = 1, -1, -2$ is actually necessary to use this method of partial fraction decomposition. Sorry, I feel very stupid right now. – The Pointer May 13 '17 at 11:54
• We all do stuff like that at least ten times a day! If we aren't making mistakes, we are not learning! – Moo May 13 '17 at 11:54

You've written:

$$9+3s = A(s+1)(s+2)+B(s-1)(s+2)+C(s+1)(s-1).$$

This equation has to be true for any $s$, so choose clever values. For instance, $s=-1$ gives us

$$6 = B(-2)(1) \implies B=-3$$

Next $s=1$:

$$12 = A(2)(3) \implies A=2.$$

Next, $s=-2$:

$$3 = C(-1)(-3) \implies C=1.$$

One can also plug in complex values for $s$ and equate real and imaginary parts.

• Thanks for the response. Why is it that we are allowed to select $s = 1, -1, -2$ when we have $\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2} \forall s \not = 1, -1, -2$? What confused me was that, initially, it would seem that we cannot select these values for $s$ (because they would result in division by $0$). – The Pointer May 13 '17 at 12:33
• When we clear the fractions, we create new functions with removable singularities. The graph of the new function looks exactly like the polynomial, except there are holes at the three singular points. But that just means the polynomials themselves have to be equal, so we work with them instead of the rational functions. – B. Goddard May 13 '17 at 12:37

You've got the right expression $$9 + 3s = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1).$$ Now you can think like this: it is a polynomial equation of the kind $p(s)=0$. A nonzero polynomial cannot have more than the finite number of zeros. Since this equation has infinitely many solutions (all $s$ except a finite number) then the polynomial must be identical zero, and the equation holds true for those exceptional values as well. Setting $s=\pm 1, -2$ we get $A=2$, $B=-3$, $C=1$.

It is also possible - as you did - to set other values of $s$ and get the linear system. In your case you have obtained $$\begin{cases} 2A-2B-C=9,\\ 12A+4B+3C=15,\\ 20A+10B+8C=18. \end{cases}$$ Using the linear algebra technique such systems are usually solved by Gauss elimination, but it is also possible to solve it by substitution as you tried, but got infinite loops. You have to do a systematic substitution: express one variable e.g. $C$ from one equation and substitute it into all other equations. Here how it goes: the first equation gives $C=2A-2B-9$. Substitute it into the second and the third ones $$\begin{cases} C=2A-2B-9,\\ 12A+4B+3(2A-2B-9)=15,\\ 20A+10B+8(2A-2B-9)=18. \end{cases}\quad\Leftrightarrow\quad \begin{cases} C=2A-2B-9,\\ 9A-B=21,\\ 18A-3B=45. \end{cases}$$ Now we do the same: express $B$ from the second equation $B=9A-21$ and set into the third one $$\begin{cases} C=2A-2B-9,\\ B=9A-21,\\ 18A-3(9A-21)=45. \end{cases}\quad\Leftrightarrow\quad \begin{cases} C=2A-2B-9,\\ B=9A-21,\\ A=2. \end{cases}$$ It works without infinite loops.

I would prefer a simpler method for this problem, known as Heaviside cover-up method: $$\dfrac{9+3s}{s^3+2s^2-s-2}=\dfrac{9+3s}{s^2(s+2)-(s+2)} =\dfrac{9+3s}{(s-1)(s+1)(s+2)}$$ Now,$$\dfrac{9+3s}{(s-1)(s+1)(s+2)}\\=\left.\dfrac{9+3s}{(s+1)(s+2)}\right|_{s=1}\cdot\dfrac{1}{s-1}+\left.\dfrac{9+3s}{(s-1)(s+2)}\right|_{s=-1}\cdot\dfrac{1}{s+1}+\left.\dfrac{9+3s}{(s-1)(s+1)}\right|_{s=-2}\cdot\dfrac{1}{s+2}\\ =\dfrac{2}{s-1}-\dfrac{3}{s+1}+\dfrac{1}{s+2}$$

• This is very neat. – Pixel May 13 '17 at 12:36

The method you are using is: if the equality holds for all values of s, then it will hold for the special ones, but that isn't enough. In the equation $$3s+9=A(s+1)(s+2)+B(s-1)(s+2) +C(s+1)(s-1)$$ multiply everything on the right side and write it in a form $$\alpha_1 s^2+\alpha_2 s +\alpha_3$$ . $\alpha_1$ is 0 because it's 0 on the left side, $\alpha_2=3, \alpha_3=9$.

I would like to mention that the reason your method usually works is that there is a theorem that says this equation always has the solution for A,B,C, otherwise you would have to check that it holds for the A,B,C you've calculated. But to get them, you need to choose "good" values for s.