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Partial fractions

Decompose algebraic fractions with distinct linear factors. Learn substitution and coefficient comparison, including non-monic factors and three-term examples, with worked practice.

Before you startFactorising polynomials, algebraic fractions and simultaneous equations

01 / What and why

Reverse the process of adding fractions.

Adding simple fractions produces a single rational expression. Partial fractions reverse that operation. The result can be easier to integrate or expand later.

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

In this lesson the numerator degree is smaller than the denominator degree, and the denominator factors into distinct linear factors. Give each linear factor its own constant numerator. Retain the excluded inputs of the original expression.

N(x)/[(x − a)(x − b)]
= A/(x − a) + B/(x − b)

Here a ≠ b, the fraction is proper, and x ≠ a, b.

Repeated factors need additional powers in the template. An improper fraction needs a polynomial part. These are separate cases, not exceptions you can ignore.

02 / Clear denominators

Create an identity before substituting excluded inputs.

Start with (5x + 1)/[(x − 1)(x + 2)] = A/(x − 1) + B/(x + 2), with x ≠ 1, −2. Multiplying by the common denominator gives:

5x + 1 ≡ A(x + 2) + B(x − 1)

The symbol ≡ means an identity: the two polynomials agree for every x. The original fractions agree only where they are defined. The polynomial identity extends to x = 1 and x = −2, so we can use those values to find A and B.

Why does the identity extend to those inputs?

Subtract the polynomials. If they agree at every allowed x, their difference has infinitely many roots. A nonzero polynomial has only finitely many roots, so the difference must be the zero polynomial. This argument does not assign a value to an undefined fraction.

Substitute the rootsWorked example

x = 1: 6 = 3A

The B term vanishes. Hence A = 2.

x = −2: −9 = −3B

The A term vanishes. Hence B = 3.

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

Keep x ≠ 1, −2. Recombine to check the numerator 2(x + 2) + 3(x − 1) = 5x + 1.

Watch root substitution isolate each coefficient

Pause, replay or seek freely. The notes explain the same idea and stay in view.

03 / Compare coefficients

Match the whole polynomial, not one sample value.

Instead of substituting roots, expand the identity. Gather the x terms and the constant terms:

5x + 1 ≡ (A + B)x + (2A − B)

Corresponding coefficients must agree. Solve A + B = 5 and 2A − B = 1. Adding gives 3A = 6, so A = 2 and B = 3.

One input generally gives only one equation for two unknowns. For example, x = 0 gives 2A − B = 1; many pairs satisfy that equation, but most fail to match the x coefficient.

Use the controls to see why both equations are necessary. This model checks the numerator identity; it does not change the original denominator or its domain.

Build the numeratorExplore

A/(x − 1) + B/(x + 2)

After taking a common denominator, the numerator is:

A(x + 2) + B(x − 1)
= (A + B)x + (2A − B)

A 2B 3

A = 2, B = 3: numerator = 5x + 1.

These values match the target 5x + 1.

Try to make the x coefficient correct while the constant is wrong. Both must match. The original restrictions x ≠ 1, −2 stay fixed for every choice, even when a numerator becomes zero.

04 / Non-monic factors

Keep the coefficient inside its factor.

A factor such as 2x − 1 still gets a constant numerator. Its root is 1/2, not 1. Do not silently replace it by x − 1/2: that would introduce a factor of 2.

Decompose (7x + 1)/[(2x − 1)(x + 1)]Worked example

A/(2x − 1) + B/(x + 1)

Original restrictions: x ≠ 1/2, −1.

7x + 1 ≡ A(x + 1) + B(2x − 1)

Each coefficient multiplies the factor missing from its own denominator.

x = 1/2: 9/2 = (3/2)A
⇒ A = 3

Substitute the exact fractional root.

x = −1: −6 = −3B
⇒ B = 2

So the answer is 3/(2x − 1) + 2/(x + 1), retaining both exclusions.

3(x + 1) + 2(2x − 1)
= 7x + 1

This independent recombination checks the result.

05 / Three factors

Each term keeps the other two factors.

For three distinct linear factors, use three constants. After clearing denominators, each coefficient multiplies the product of the other two factors.

Decompose (2x² + 6x − 2)/[x(x − 1)(x + 2)]Worked example

A/x + B/(x − 1) + C/(x + 2)

x ≠ 0, 1, −2.

2x² + 6x − 2
≡ A(x − 1)(x + 2)
+ Bx(x + 2) + Cx(x − 1)

Substituting a root kills two terms at once.

x = 0: −2 = −2A ⇒ A = 1
x = 1: 6 = 3B ⇒ B = 2
x = −2: −6 = 6C ⇒ C = −1

Keep signs when evaluating the remaining product.

Answer: 1/x + 2/(x − 1)
− 1/(x + 2)

Recombination gives (x² + x − 2) + (2x² + 4x) − (x² − x) = 2x² + 6x − 2.

06 / Factor the denominator

The expanded form can hide the template.

Factor the denominator before deciding how many fractions you need. A quadratic with two distinct real linear factors uses the same two-constant method. For a cubic, extract a common factor or use a known root first.

A factorisable quadraticWorked example

(x + 8)/(x² + x − 6)

x² + x − 6 = (x − 2)(x + 3); x ≠ 2, −3.

x + 8 ≡ A(x + 3) + B(x − 2)

At x = 2, 10 = 5A, so A = 2. At x = −3, 5 = −5B, so B = −1.

= 2/(x − 2) − 1/(x + 3)

Check: 2(x + 3) − (x − 2) = x + 8.

Check the template first

(x − 2)(x + 3): two distinct linear factors.
x(x − 2)(x + 3): three distinct linear factors.
(x − 2)²(x + 3): a repeated factor; this lesson’s simple template is insufficient.
x² + 1: no real linear factors; the current method does not apply directly.

The first check is always whether the fraction is proper: numerator degree must be smaller than denominator degree.

07 / Check your answer

A zero coefficient is possible.

Recombine over the original denominator. Compare the resulting numerator with the original one and retain its restrictions. Coefficients may be positive, negative, fractional or zero.

(6x + 12)/[(x + 2)(x − 4)]
= 0/(x + 2) + 6/(x − 4)
= 6/(x − 4), x ≠ −2, 4

The zero coefficient makes one term disappear. It does not put x = −2 back into the original domain.

Why may x = 1 be used to find A in a decomposition containing A/(x − 1)?

Hint

Distinguish the rational expression from the polynomial identity.

Worked solution

You do not evaluate A/(x − 1) at 1. First clear denominators to obtain a polynomial identity. That identity holds at 1 and isolates A. The original fraction remains undefined at 1.

09 / Recap

Template, identity, coefficients, check.

Check the fraction is proper and factor its denominator.
Assign one constant numerator to each distinct linear factor.
Clear denominators to make a polynomial identity.
Substitute factor roots or compare coefficients.
Recombine and retain the original excluded inputs.

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Section 1 of 9 · What and why