Partial Fraction Coefficients

Oct 2012
1,314
21
USA
What is the best strategy for finding coefficients, given the following scenarios:

A. Distinct linear factors. Ex: \(\displaystyle (x + 3) (x + 2)\) etc..

Equation System, or Substitution method

B. Repeated Linear Factors Ex \(\displaystyle (x + 3)^{2}\) or \(\displaystyle x^{2}\)

?

C. Irreducible Quadratic \(\displaystyle (x^{2} + 25)\)

Equation System only ??

Can we use an equation system in all cases? I think not, as was not the case in one problem.
 
Last edited:
Feb 2015
172
39
Upstate NY
This information is in your textbook and online but here goes:

A) write 1/[(x+3)(x+2) as [A/(x+3)] + [B/(x+2)]

B) write 1/(x + 2)^3 as A/(x+2)^3 + B/(x+2)^2 + C/(x+2)

C) write 1/(x^2+25) as (Ax+B)/(x^2+25)
 
Oct 2012
1,314
21
USA
What is the best strategy for finding the actual A, B C values not just the setup.
 
Feb 2015
172
39
Upstate NY
What is the best strategy for finding the actual A, B C values not just the setup.
for what I wrote on the right hand sides you need to get a common denominator, which happens to be the same denominator as the left hand side equations. Then just equate the numerators and solve for the unknowns.
 
Oct 2012
1,314
21
USA
for what I wrote on the right hand sides you need to get a common denominator, which happens to be the same denominator as the left hand side equations. Then just equate the numerators and solve for the unknowns.
That's true, but not for irreducible quadratics, and maybe not for repeated linear factors.
 

Ted

Feb 2010
240
64
China
Distinct linear factors ---> Substitution method is better.
As each factor zero will give value of a constant directly.
Ex :
\(\displaystyle \frac{1}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}\)

\(\displaystyle 1 = A (x+2)(x+3) + B (x+1)(x+3) + C (x+1)(x+2)\)

Note that putting x=-1 will give the value of A directly, x=-2 will give B & x=-3 will give C.

If there is quadratic factors, equating method is better.