Question

With partial fractions, if the numerator is a constant, will A, B, etc be that constant...with linear denominator partials...not nonlinear denominators like this problem: http://math.stackexchange.com/questions/130633/partial-fraction-with-a-constant-as-numerator)

Answer 1

\[\frac{A}{(x-c_1)(x-c_2)} \text{ can be rewritten using partial fractions } \\ \\ \text{ both of the factors in the denominator are linear } \\ \text{ so we can write } \\ \frac{A}{(x-c_1)(x-c_2)} \text{ as } \frac{B}{x-c_1}+\frac{D}{x-c_2} \\ \text{ Now we need to find } \\ B \text{ and } D \text{ in terms of } A ,c_2,c_1 \\ \text{ Let's combine the fractions now } \\ \frac{B}{x-c_1}+\frac{D}{x-c_2}=\frac{Bx-Bc_2+Dx-Dc_1}{(x-c_1)(x-c_2)} =\frac{(B+D)x-Bc_2-Dc_1}{(x-c_1)(x-c_2)} \\ \text{ this implies } \\ B+D=0 \text{ and } -Bc_2-Dc_1=A \\ \text{ so } B=-D \\ \text{ we will use this into the second equation } \\ -Bc_2-Dc_1=A \\ -Bc_2+Bc_1=A \\ \text{ now solving for } B \\ \\ B(-c_2+c_1)=A \\ B=\frac{A}{-c_2+c_1} \\ D=-B=\frac{-A}{-c_2+c_1}=\frac{A}{c_2-c_1}\]

\[\text{ So you can write your fraction } \\ \frac{A}{(x-c_1)(x-c_2)} \text{ as } \frac{\frac{A}{-c_1+c_2}}{x-c_1}+\frac{\frac{A}{c_2-c_1}}{x-c_2}\]

anyways that is just an example
first step make sure deg of bottom is greater than deg of top
then if your factors are linear on bottom then your fractions in your sum will have constant numerators