Algebra · Quick Reference

Partial Fractions Cheat Sheet

Algebra II · Lesson 17/17

In one line: partial fraction decomposition breaks a complicated rational expression back down into a sum of simpler fractions — a technique that makes many integrals and equations far easier to handle.

Key Ideas

1What Partial Fractions Do. Rewrite a single complex fraction, like (5x+1)/((x+1)(x-2)), as a sum of simpler fractions with smaller denominators.

2Setting Up the Decomposition. For distinct linear factors, assign one unknown constant to each factor's fraction: (5x+1)/((x+1)(x-2)) = A/(x+1) + B/(x-2).

3Solving for the Unknown Constants. Multiply both sides by the original denominator to clear fractions, then substitute convenient x-values (or match coefficients) to solve for each constant.

4Repeated Linear Factors. A repeated factor like (x-1)^2 requires a term for each power: A/(x-1) + B/(x-1)^2.

5Why It's Useful. Simpler fractions are much easier to integrate, differentiate, or otherwise manipulate than one complicated combined fraction.

Worked Examples

Decompose (5x+1) / [(x+1)(x-2)] into partial fractions.

(5x+1)/[(x+1)(x-2)] = (4/3)/(x+1) + (11/3)/(x-2)

Decompose (3x+5) / [(x-1)(x+3)] into partial fractions.

(3x+5)/[(x-1)(x+3)] = 2/(x-1) + 1/(x+3)

Check the decomposition 2/(x-1) + 1/(x+3) by combining back into a single fraction and confirming the numerator.

Confirmed correct

Formulas

(px+q) / [(x-a)(x-b)] = A/(x-a) + B/(x-b)

Practice Yourself

Set up the partial fraction form for (2x+3)/[(x-1)(x+2)].

A/(x-1) + B/(x+2)

For 5x/[(x-2)(x+1)] = A/(x-2)+B/(x+1), find A by letting x=2.

5(2)=A(2+1), 10=3A, A=10/3

What kind of denominator does partial fraction decomposition break apart?

A product of simpler factors, usually linear