Math Core

You already know how to add fractions like $\dfrac{2}{x - 1} - \dfrac{1}{x + 3}$ into a single fraction. Partial fraction decomposition runs that process backward: it takes one complicated rational expression and splits it into a sum of simpler ones. It is a natural end to this unit because finding the pieces comes down to solving a system of linear equations, and it is a tool you will use constantly in calculus, where simple fractions are far easier to work with than complicated ones.

The idea

Start with a sum you can compute:

\frac{2}{x - 1} - \frac{1}{x + 3} = \frac{2(x + 3) - (x - 1)}{(x - 1)(x + 3)} = \frac{x + 7}{(x - 1)(x + 3)}.

Now suppose you were handed only the right side. Because its denominator factors into $(x - 1)$ and $(x + 3)$ , you can guess that it came from fractions with those denominators, and write

\frac{x + 7}{(x - 1)(x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 3}

for some constants $A$ and $B$ . The job is to find them.

Step 0: make sure the fraction is proper

Partial fractions work directly only on a proper rational expression, one where the degree of the numerator is less than the degree of the denominator. If it is not, divide first (long division or synthetic division) to write it as a polynomial plus a proper fraction, then decompose the proper part.

For example, $\dfrac{x^2 + 1}{x^2 - 1}$ has equal degrees, so divide: $\dfrac{x^2 + 1}{x^2 - 1} = 1 + \dfrac{2}{x^2 - 1}$ . Only $\dfrac{2}{x^2 - 1}$ gets decomposed.

The forms to use

Factor the denominator completely into linear factors and irreducible quadratic factors (quadratics like $x^2 + 4$ that have no real zeros). Each factor contributes its own terms.

Partial fraction forms

Each distinct linear factor $(ax + b)$ contributes $\dfrac{A}{ax + b}$ .
A repeated linear factor $(ax + b)^k$ contributes one term for each power up to $k$ : $\dfrac{A_1}{ax + b} + \dfrac{A_2}{(ax + b)^2} + \cdots + \dfrac{A_k}{(ax + b)^k}$ .
Each irreducible quadratic factor $(ax^2 + bx + c)$ contributes $\dfrac{Bx + C}{ax^2 + bx + c}$ , with a linear numerator.

For example, $\dfrac{x + 3}{(x - 1)(x^2 + 4)}$ has the form $\dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 4}$ , and $\dfrac{5}{x(x + 2)^3}$ has the form $\dfrac{A}{x} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^2} + \dfrac{D}{(x + 2)^3}$ .

Finding the constants

Multiply both sides by the full denominator to clear fractions. You get a polynomial identity that must hold for every $x$ . Then use either (or both) of these methods:

Substitute convenient values. Plug in the zeros of the linear factors; each one wipes out every term but one.
Match coefficients. Expand, collect like powers of $x$ , and set coefficients on each side equal. This gives a system of linear equations in the unknown constants, exactly the kind of system from the start of this unit.

Worked example: Distinct linear factors

Decompose $\dfrac{x + 7}{(x - 1)(x + 3)}$ .

Write $\dfrac{x + 7}{(x - 1)(x + 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 3}$ and multiply by $(x - 1)(x + 3)$ :

x + 7 = A(x + 3) + B(x - 1).

Set $x = 1$ : $8 = 4A$ , so $A = 2$ . Set $x = -3$ : $4 = -4B$ , so $B = -1$ . Therefore

\frac{x + 7}{(x - 1)(x + 3)} = \frac{2}{x - 1} - \frac{1}{x + 3},

which matches the sum we started with. Matching coefficients works too: $x + 7 = (A + B)x + (3A - B)$ gives the system $A + B = 1$ and $3A - B = 7$ , whose solution is again $A = 2$ , $B = -1$ .

Worked example: Factor first

Decompose $\dfrac{3x + 11}{x^2 - x - 6}$ .

The denominator factors as $(x - 3)(x + 2)$ . Write $3x + 11 = A(x + 2) + B(x - 3)$ .

Set $x = 3$ : $20 = 5A$ , so $A = 4$ . Set $x = -2$ : $5 = -5B$ , so $B = -1$ .

\frac{3x + 11}{x^2 - x - 6} = \frac{4}{x - 3} - \frac{1}{x + 2}.

Worked example: A repeated linear factor

Decompose $\dfrac{x^2 + 2}{x(x - 1)^2}$ .

The factor $(x - 1)$ appears twice, so it needs two terms:

\frac{x^2 + 2}{x(x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}.

Multiply by $x(x - 1)^2$ :

x^2 + 2 = A(x - 1)^2 + Bx(x - 1) + Cx.

Set $x = 0$ : $2 = A$ . Set $x = 1$ : $3 = C$ . No value of $x$ isolates $B$ , so compare the $x^2$ coefficients: on the left it is $1$ , and on the right it is $A + B$ . So $A + B = 1$ , giving $B = -1$ .

\frac{x^2 + 2}{x(x - 1)^2} = \frac{2}{x} - \frac{1}{x - 1} + \frac{3}{(x - 1)^2}.

Worked example: An irreducible quadratic factor

Decompose $\dfrac{3x^2 - x - 5}{(x - 2)(x^2 + 1)}$ .

Since $x^2 + 1$ has no real zeros, its numerator must be linear:

3x^2 - x - 5 = A(x^2 + 1) + (Bx + C)(x - 2).

Set $x = 2$ : $12 - 2 - 5 = 5A$ , so $A = 1$ . Expand the right side with $A = 1$ :

x^2 + 1 + Bx^2 - 2Bx + Cx - 2C = (1 + B)x^2 + (C - 2B)x + (1 - 2C).

Matching: $1 + B = 3$ gives $B = 2$ , and $1 - 2C = -5$ gives $C = 3$ . Check the middle coefficient: $C - 2B = 3 - 4 = -1$ . ✓

\frac{3x^2 - x - 5}{(x - 2)(x^2 + 1)} = \frac{1}{x - 2} + \frac{2x + 3}{x^2 + 1}.

Common mistake

Two setups cause most wrong answers. First, a repeated factor $(x - 1)^2$ needs both $\dfrac{B}{x - 1}$ and $\dfrac{C}{(x - 1)^2}$ ; leaving out the lower power makes the system impossible to solve. Second, an irreducible quadratic needs the linear numerator $Bx + C$ , not just a constant.

Tip

Check your answer by plugging in a value of $x$ that you did not use, such as $x = 0$ or $x = 5$ , into both the original expression and your decomposition. In the first example at $x = 0$ : the original is $\dfrac{7}{-3}$ , and the decomposition gives $\dfrac{2}{-1} - \dfrac{1}{3} = -\dfrac{7}{3}$ . ✓

Practice

Practice 1

Find the constants $A$ and $B$ so that $\dfrac{5x + 1}{(x - 1)(x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}$ . Give your answer as $(A, B)$ .

Enter a point like (2, -3)

Practice 2

Find $A$ and $B$ so that $\dfrac{6}{x^2 - 9} = \dfrac{A}{x - 3} + \dfrac{B}{x + 3}$ . Give your answer as $(A, B)$ .

Enter a point like (2, -3)

Practice 3

Write $\dfrac{4x + 1}{x^2 - x - 2}$ as $\dfrac{A}{x - 2} + \dfrac{B}{x + 1}$ . Give $(A, B)$ .

Enter a point like (2, -3)

Practice 4

Find $A$ and $B$ so that $\dfrac{3x - 1}{(x - 2)^2} = \dfrac{A}{x - 2} + \dfrac{B}{(x - 2)^2}$ . Give $(A, B)$ .

Enter a point like (2, -3)

Practice 5

Which is the correct form of the partial fraction decomposition of $\dfrac{x + 3}{(x - 1)(x^2 + 4)}$ ?

Practice 6

The expression $\dfrac{2x^2 - 3x - 1}{x^3 - x}$ can be written as $\dfrac{A}{x} + \dfrac{B}{x + 1} + \dfrac{C}{x - 1}$ . Find $B$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Find $A$ , $B$ and $C$ so that $\dfrac{x^2 + x + 4}{x(x^2 + 2)} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 + 2}$ . Give $(A, B, C)$ .

Enter a point like (2, -3)

Practice 8

Divide first, then decompose: $\dfrac{2x^2 + x - 5}{x^2 - 1} = 2 + \dfrac{A}{x - 1} + \dfrac{B}{x + 1}$ . Find $(A, B)$ .

Enter a point like (2, -3)