Algebraic Fractions and Partial Fractions

This lesson covers simplifying algebraic fractions, performing arithmetic with them, and decomposing rational expressions into partial fractions — a technique essential for integration at A-Level.

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Simplifying Algebraic Fractions

To simplify an algebraic fraction, factorise the numerator and denominator, then cancel common factors.

Worked Example 1

Simplify (x² − 9)/(x² + 5x + 6).

Numerator: x² − 9 = (x − 3)(x + 3)
Denominator: x² + 5x + 6 = (x + 2)(x + 3)

Cancel (x + 3):

= (x − 3)/(x + 2), provided x ≠ −3

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Adding and Subtracting Algebraic Fractions

Find a common denominator, combine, and simplify.

Worked Example 2

Express 3/(x + 1) − 2/(x − 3) as a single fraction.

Common denominator: (x + 1)(x − 3)

= [3(x − 3) − 2(x + 1)] / [(x + 1)(x − 3)]
= [3x − 9 − 2x − 2] / [(x + 1)(x − 3)]
= (x − 11) / [(x + 1)(x − 3)]

Answer: (x − 11) / [(x + 1)(x − 3)]

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Multiplying and Dividing Algebraic Fractions

Multiply: multiply numerators together and denominators together, then simplify.

Divide: multiply by the reciprocal.

Worked Example 3

Simplify [(x² − 4)/(x + 5)] × [(x + 5)/(x + 2)].

= [(x − 2)(x + 2)/(x + 5)] × [(x + 5)/(x + 2)]

Cancel (x + 5) and (x + 2):

= x − 2

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Improper Algebraic Fractions

An algebraic fraction is improper when the degree of the numerator is greater than or equal to the degree of the denominator. Use polynomial long division to express it as a polynomial plus a proper fraction.

Worked Example 4

Express (x³ + 2x² − x + 3)/(x + 1) in the form ax² + bx + c + d/(x + 1).

Perform polynomial division:

x³ + 2x² − x + 3 = (x + 1)(x² + x − 2) + 5

So: (x³ + 2x² − x + 3)/(x + 1) = x² + x − 2 + 5/(x + 1)

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Partial Fractions

Partial fractions decompose a proper fraction into simpler fractions. This is essential for integration and series work.

Type 1: Linear Factors

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

Worked Example 5

Express (5x + 3)/[(x + 1)(x − 2)] in partial fractions.

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

Multiply through: 5x + 3 = A(x − 2) + B(x + 1)

Set x = 2: 13 = 3B → B = 13/3
Set x = −1: −2 = −3A → A = 2/3

= 2/[3(x + 1)] + 13/[3(x − 2)]

Type 2: Repeated Linear Factor

f(x)/[(x − a)²(x − b)] = A/(x − a) + B/(x − a)² + C/(x − b)

Worked Example 6

Express (3x² + x − 2)/[(x − 1)²(x + 2)] in partial fractions.

Let = A/(x − 1) + B/(x − 1)² + C/(x + 2)

Multiply: 3x² + x − 2 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)²

Set x = 1: 3 + 1 − 2 = 3B → B = 2/3
Set x = −2: 12 − 2 − 2 = 9C → C = 8/9
Compare x² coefficients: 3 = A + C → A = 3 − 8/9 = 19/9

= 19/[9(x − 1)] + 2/[3(x − 1)²] + 8/[9(x + 2)]

Type 3: Irreducible Quadratic Factor

f(x)/[(x − a)(x² + bx + c)] = A/(x − a) + (Bx + C)/(x² + bx + c)

Worked Example 7

Express (4x² + 1)/[(x − 1)(x² + 1)] in partial fractions.

Let = A/(x − 1) + (Bx + C)/(x² + 1)

Multiply: 4x² + 1 = A(x² + 1) + (Bx + C)(x − 1)

Set x = 1: 5 = 2A → A = 5/2
Compare x² coefficients: 4 = A + B → B = 4 − 5/2 = 3/2
Compare constants: 1 = A − C → C = 5/2 − 1 = 3/2

= 5/[2(x − 1)] + (3x + 3)/[2(x² + 1)]

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Exam Tips

• Always factorise fully before simplifying or decomposing.
• For partial fractions, the cover-up method (substituting roots) is quick for linear factors.
• Check your answer by recombining the partial fractions.
• Identify the type of denominator first: distinct linear, repeated linear, or irreducible quadratic.
• Improper fractions must be divided out before applying partial fractions.