Partial Fraction – SS3 Further Mathematics Lesson Note (First Term)

Partial fractions are simpler fractions that add up to one fraction with a factorised denominator.

Behavioural objectives

By the end of the lesson, students should be able to:

1. state when long division is needed before resolving a fraction into partial fractions;
2. divide the numerator by the denominator to get a quotient and a remainder;
3. write the remainder fraction as partial fractions with constants A and B;
4. find A and B by substituting suitable values of x;
5. check an answer by equating coefficients.

What are partial fractions?

Partial fractions are the simpler fractions obtained when one fraction is split according to the factors of its denominator.

• Example: 9/(x + 1) − 12/(x + 2) is a pair of partial fractions.
• Use: they make integration, series expansion and algebraic simplification easier.
• Requirement: the denominator must first be written in factors, such as (x + 1)(x + 2).

What is the difference between a proper and an improper fraction in algebra?

A proper algebraic fraction has a numerator of lower degree than its denominator. An improper one has a numerator of equal or higher degree.

When is long division needed in partial fractions?

Long division is needed when the degree of the numerator is equal to or higher than the degree of the denominator.

• Quotient: the whole-number or polynomial part, such as 2 or 3x + 16.
• Remainder: the left-over part, which has a lower degree than the denominator.
• Next step: resolve only the remainder fraction into partial fractions.

What are the forms of partial fractions?

The form of the partial fractions depends on the type of factor in the denominator.

• Note: the worked examples below use two different linear factors.

How do you resolve an improper fraction into partial fractions?

Divide first, then split the remainder fraction, then find the constants.

1. Factorise the denominator, or expand it if you need it for the division.
2. Compare the degrees of the numerator and the denominator.
3. If the numerator degree is equal or higher, divide to get a quotient and a remainder.
4. Write the remainder fraction as A/(first factor) + B/(second factor).
5. Multiply both sides by the denominator to remove the fractions.
6. Put in values of x that make one bracket zero, to find A and B.
7. Write the final answer as the quotient plus the partial fractions.

Worked examples

Example 1

Resolve (2x² + 3x + 10) / ((x + 1)(x + 2)) into partial fractions.

Solution:

1. Expand the denominator: (x + 1)(x + 2) = x² + 3x + 2.
2. The degrees are equal (both 2), so divide. The quotient is 2.
3. 2 × (x² + 3x + 2) = 2x² + 6x + 4.
4. Subtract: 2x² + 3x + 10 − (2x² + 6x + 4) = −3x + 6. This is the remainder.
5. So (2x² + 3x + 10) / ((x + 1)(x + 2)) = 2 + (−3x + 6) / ((x + 1)(x + 2)).
6. Let (−3x + 6) / ((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2).
7. Then −3x + 6 = A(x + 2) + B(x + 1).

• Put x = −2: 12 = −B, so B = −12.
• Put x = −1: 9 = A, so A = 9.

Hence (−3x + 6) / ((x + 1)(x + 2)) = 9/(x + 1) − 12/(x + 2).

Answer: (2x² + 3x + 10) / ((x + 1)(x + 2)) = 2 + 9/(x + 1) − 12/(x + 2).

• Check: A(x + 2) + B(x + 1) = (A + B)x + (2A + B).
• Coefficient of x: 9 − 12 = −3. This is correct.
• Constant term: 18 − 12 = 6. This is correct.

Example 2

Resolve (3x³ + 22x² + 38x + 7) / ((x − 1)(x + 3)) into partial fractions.

Solution:

1. Expand the denominator: (x − 1)(x + 3) = x² + 2x − 3.
2. The numerator has degree 3 and the denominator has degree 2, so divide.
3. 3x × (x² + 2x − 3) = 3x³ + 6x² − 9x. Subtracting leaves 16x² + 47x + 7.
4. 16 × (x² + 2x − 3) = 16x² + 32x − 48. Subtracting leaves 15x + 55.
5. The quotient is 3x + 16 and the remainder is 15x + 55.
6. So the fraction = 3x + 16 + (15x + 55) / ((x − 1)(x + 3)).
7. Let (15x + 55) / ((x − 1)(x + 3)) = A/(x − 1) + B/(x + 3).
8. Then 15x + 55 = A(x + 3) + B(x − 1).

• Put x = −3: 10 = −4B, so B = −5/2.
• Put x = 1: 70 = 4A, so A = 35/2.

Answer: (3x³ + 22x² + 38x + 7) / ((x − 1)(x + 3)) = 3x + 16 + 35/(2(x − 1)) − 5/(2(x + 3)).

• Check: A(x + 3) + B(x − 1) = (A + B)x + (3A − B).
• Coefficient of x: 35/2 − 5/2 = 15. This is correct.
• Constant term: 105/2 + 5/2 = 55. This is correct.

What mistakes should students avoid?

• Skipping the division: never split an improper fraction directly.
• Sign errors: subtract the whole line, so 3x − 6x gives −3x.
• Wrong value of x: choose the value that makes a bracket equal to zero.
• Leaving out the quotient: the final answer must include it.
• No check: compare coefficients of x and the constant terms.

Evaluation

1. When must you use long division before resolving a fraction into partial fractions?
2. In Example 1, what are the quotient and the remainder after the division?
3. In Example 1, why is x = −2 a useful value to put into −3x + 6 = A(x + 2) + B(x + 1)?
4. In Example 2, state the values of A and B.

Assignment

Resolve into partial fractions:

1. x² / ((x − 1)(x + 3))
2. (x³ − 1) / (x² − 4)
3. (2x² + 3x + 1) / (x(x + 1))

References

1. Multipurpose Mathematics for SSCE, NECO and GCE, 2nd Edition, by J. Olowofero.
2. New Mathematics Project 3 by M.R. Tuttuh, Adegun et al.