# Partial Fraction Coefficients

#### Jason76

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.

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#### JaguarXJS

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)

#### Jason76

What is the best strategy for finding the actual A, B C values not just the setup.

#### JaguarXJS

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.

#### Jason76

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

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.