Partial Fractions

Adding and over a common denominator produces . Partial fraction decomposition is that process in reverse: given the messy single fraction, recover the two simple ones. The pieces are easier to work with in almost every setting that matters — series, transforms, and integration all prefer them.

What Is a Partial Fraction?

A partial fraction is one of the simple pieces a complicated rational expression breaks into — a single term whose denominator is one factor of the original denominator. Finding those pieces is called partial fractions decomposition, and it is the only thing this lesson does.

and are the partial fractions of .

Nothing is approximated. The sum equals the original for every value of where both sides are defined — it is the same expression, written in a more useful shape.

The Degree Check Comes First

Partial fractions only apply to a proper rational expression — one whose numerator has a strictly lower degree than its denominator.

If it is not proper, run polynomial long division first. The quotient stays as it is and only the remainder fraction gets decomposed.

Example of the fix. For , division gives:

Now the leftover piece is proper, and only needs splitting.

Partial Fraction Decomposition Rules

Factor the denominator completely, then read one rule per factor. Every factor of the denominator contributes its own term or terms to the sum.

A repeated factor needs every power, not just the highest. Writing only for throws away a degree of freedom, and the constants will not solve.

“Irreducible” means irreducible over the real numbers. Check the discriminant: has , so it stays whole. But has discriminant , so it must be split into before you write any terms — see factoring polynomials if the factoring itself is the sticking point.

The Method in Four Steps

For step 4 there are two ways to finish, and they can be mixed freely:

• Substitute convenient values. Each root of the denominator kills all but one constant, giving it in a single line. This is fastest for distinct linear factors.
• Compare coefficients. Expand both sides and match the coefficient of each power of . This is what you need when a quadratic factor leaves constants that no substitution isolates.

Worked Example: Two Distinct Linear Factors

Decompose .

Step 1 — degree check. The numerator is degree , the denominator degree , so the fraction is proper.

Step 2 — the denominator is already factored, into two distinct linear factors.

Step 3 — write the form, one constant per factor:

Step 4 — multiply through by to clear the denominators:

Substitute , which kills the term:

Substitute , which kills the term:

Always check by recombining. Here , which matches the original numerator. That one line catches almost every arithmetic slip.

Worked Example: A Repeated Linear Factor

Decompose .

Step 1 — the form gets a term for each power of the repeated factor:

Step 2 — clear denominators by multiplying through by :

Step 3 — substitute :

Step 4 — compare the coefficients of . On the left it is ; on the right it is :

Only one substitution was available here, because the denominator has a single root. Coefficient comparison finished the job.

Worked Example: An Irreducible Quadratic Factor

Decompose .

Step 1 — has discriminant , so it is irreducible and takes a linear numerator:

Step 2 — clear denominators:

Step 3 — substitute , the only real root available:

Step 4 — expand the right side and collect powers:

The constant row repeats what substitution already gave, which is exactly the redundancy you want — it confirms the other two rows.

Common Mistakes to Avoid

• Skipping the degree check. An improper fraction produces an inconsistent system, and the usual reaction is to hunt for an arithmetic error that is not there.
• Giving a repeated factor one term. needs three terms, with denominators , and .
• Putting a constant over a quadratic. An irreducible quadratic takes on top, not . One constant is not enough to match both a linear and a constant term.
• Leaving a factorable quadratic whole. looks like a quadratic factor, but it is and belongs in the linear rules.
• Forgetting to factor the denominator completely. The rules apply to fully factored denominators; a half-factored one gives a form that cannot be solved.
• Not checking the answer. Recombining the pieces takes one line and confirms every constant at once.

Practice Problems

Work each before opening the answer.

Problem 1. Decompose .

Problem 2. Write the decomposition form for . Do not solve for the constants.

Problem 3. Why can not be decomposed as it stands?

Problem 4. Decompose .

Problem 5. Decompose .

Problem 6. A student treats as an irreducible quadratic factor. What went wrong?

Decomposition is where the algebra of this chapter meets the rest of mathematics: the pieces it produces are exactly the forms that rational functions are analysed through, and the same splitting turns an otherwise intractable integral into a sum of routine ones. Everything it relies on — factoring, long division, and comfort with rational expressions — is already in this chapter.