This document outlines a shortcut for partial fractions involving 2 factors in the denominator (P(x) + a) and (P(x) + b).
At the end of the document it gives a challenge to find a similar shortcut for 3 factors (P(x) + a), (P(x) + b) and (P(x) + c). I have not been able to come up with anything. Was wondering if anyone else can figure this one out.
On
it is the term $\frac{1}{(p(x)+a)(p(x)+b)(p(x)+c)}$ the solution is given by ${\frac {1}{ \left( -c+a \right) \left( -c+b \right) \left( p(x)+c \right) }}-{\frac {1}{ \left( -b+a \right) \left( -c+b \right) \left( p(x)+b \right) }}+{\frac {1}{ \left( -c+a \right) \left( -b+a \right) \left( p(x)+a \right) }}$
It is exactly the same philosophy as for two factors; for simplicity, I shall note $p$ instead of $p(x)$.
Write $$\frac{1}{(p+a)(p+b)(p+c)}=\frac{A}{p+a}+\frac{B}{p+b}+\frac{C}{p+c}$$ Reduce to same denominator. Then$$1=A(p+b)(p+c)+B(p+a)(p+c)+C(p+a)(p+b)$$ Now, make $p=-a$ and get $$1=A(b-a)(c-a)$$ Do the same with $p=-b$ and get $$1=B(a-b)(c-b)$$ and again for $p=-c$ to get $$1=C(a-c)(b-c)$$ and you are done.