# Partial Fraction decomposition when denominator is in $x^2 + a$ form

Why isn't the expansion of $$\frac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)}$$

in the form of

$$\frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+16}$$ since (as I recall) the denominator is of square power and you should decompose it (in the numerator) until the s diminishes.

Apparently wolframalpha is saying the correct form should be

$$\frac{A}{s^2+1} + \frac{B}{s^2+16}$$

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You are perfectly right, that is the general form. For your particular numerator, one cannot find $A$ and $B$ such that the thing is $\frac{A}{s^2+1}+\frac{B}{s^2+16}$. – André Nicolas Aug 29 '13 at 3:57
which one, the As+B/s^2 + 1 .. is right? – 40Plot Aug 29 '13 at 3:58
Yes, the one with $\frac{As+B}{s^2+1}$. But if we use complex numbers, we can do it as $\frac{A}{s+i}+\frac{B}{s-i}+\frac{C}{s+4i}+\frac{D}{s-4i}$. – André Nicolas Aug 29 '13 at 4:00

We have:

$$\dfrac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} = \dfrac{as+b}{s^2+1} + \dfrac{cs+d}{s^2+16}$$

When we multiply things out and equate sides, we have:

$$s^3 - 2s^2 + 16s - 2 = 16 b + d + 16 a s + c s + b s^2 + d s^2 + a s^3 + c s^3$$

$$a+c = 1, b+d = -2, 16a+c = 16, 16b+d=-2$$

When you solve for the constants, you get:

$$a = 1, b = 0, c = 0, d = -2$$

So,

$$\dfrac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} = \dfrac{s}{s^2+1} - \dfrac{2}{s^2+16}$$

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how did you solve the constants so quickly – 40Plot Aug 29 '13 at 4:06
It is just algebra and simplification. By the way, I checked on WA and it agrees with the result. Are you familiar with the cover up method, but just algebra. – Amzoti Aug 29 '13 at 4:07
I tried solving it, by plugging in values - ended up with 4 equations which should give the answer once solved. But it looked tedious and messy so I thought my solution is incorrect... but apparently not. – 40Plot Aug 29 '13 at 4:10
You get a 4x4 to solve. I can add those details if you'd like, but sounds like you nailed it! I added those details for you to compare. Regards – Amzoti Aug 29 '13 at 4:13
+1 As always, clear and complete. :-) – Babak S. Aug 29 '13 at 7:16

There is a fast way, which is seldom seen on a calculus textbook. Polynomials $s^2 + 1$ and $s^2 + 16$ are coprime and squarefree, so we can solve it in a manner similar to linear denominators.

The numerator on $s^2 + 1$ is the unique representative of $$\frac {s^3 - 2s^2 + 16s - 2}{s^2 + 16} + \left\langle s^2 + 1 \right\rangle$$ whose degree less than $\deg \left( s^2 + 1 \right) = 2$. Let's denote this as $$\left\lceil s^2 + 1 \right\rceil \frac {s^3 - 2s^2 + 16s - 2}{s^2 + 16}.$$ To extract multiples of $s^2 + 1$, we can simply set $s^2 = -1$ and let the problem solves itself. $$\left\lceil s^2 + 1 \right\rceil \frac {s^3 - 2s^2 + 16s - 2}{s^2 + 16} = \frac {-\left(s - 2\right) + 16s - 2}{15} = s.$$ Similarly, $$\left\lceil s^2 + 16 \right\rceil \frac {s^3 - 2s^2 + 16s - 2}{s^2 + 1} = \frac {-16\left(s - 2\right) + 16s - 2}{-15} = -2.$$ Both numerators are found. $$\frac {s^3 - 2s^2 + 16s - 2}{\left(s^2 + 1\right) \left(s^2 + 16\right)} = \frac{s}{s^2+1} - \frac{2}{s^2+16}.$$

This method was published in Guoce Xin's A Fast Algorithm for Partial Fraction Decompositions (2004). It is covered in the last section in the article.

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