Partial Fraction Decomposition

1. For any linear factor in the denominator, the decomposition will contain a term of the form , for some constant
2. For any irreducible quadratic factor in the denominator, the decomposition will contain the term of the form , for some constants and
3. For any factor which is repeated times, we need terms of the forms given by Rules 1 and 2, but distinguished by the exponents 1 through

Only rule 1 is needed.

To determine the values of and , the idea is to put these expressions over a common denominator again and match up the coefficients

Now, equate the numerators

The only way for these polynomials to be equal is if the coefficients are equal, so we have and . Solving this linear system, we find that and , we we have our result:

Alternatively, for , we can take a shortcut

Both rules 1 and 2 are needed

Equating the numerators

By setting , we find that , so .

Then, to find the other constants:

We find that and (because there is no or term on the left side), and since , then and .

Since the fraction is improper, the top needs to be factored. We need to use polynomial long division and not synthetic division because the denominator is not linear.

So the polynomial long division of divided by is with the remainder of

• Setting gives

• Setting gives , and since ,

• Comparing terms tells us that

• Comparing terms tlls us that , and since ,

• Since , and ,