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Partial fractions: Class 10 Mathematics
Rational fractions, proper and improper forms, and splitting a fraction into partial fractions when the denominator has distinct linear factors, repeated linear factors, or an irreducible quadratic factor.
Key points
A rational fraction is a fraction N(x)/D(x) in which the numerator N(x) and the denominator D(x) are polynomials and D(x) is not zero. Examples are (x + 1)/(x² − 4) and (x³ + 2)/(x − 1).
A rational fraction is proper if the degree of the numerator is less than the degree of the denominator. It is improper if the degree of the numerator is equal to or greater than the degree of the denominator. For example, (2x + 1)/(x² − 1) is proper, while (x² + 1)/(x² − 1) is improper.
Adding fractions joins them into one fraction. For example 1/(x − 1) + 1/(x + 1) = 2x/(x² − 1). Resolving into partial fractions is the reverse: a single fraction is split into a sum of simpler fractions. The simpler fractions are called partial fractions. The method is used in algebra and later in integration.
An improper fraction must first be divided. Long division gives a quotient polynomial and a remainder, so N(x)/D(x) = Q(x) + R(x)/D(x), where R(x)/D(x) is proper. Only the proper part is then split into partial fractions.
The form of the partial fractions depends on the factors of the denominator. Factorise the denominator completely first. There are three cases, which a student should be able to recognise at once.
Case 1: distinct linear factors. For (px + q)/((x − a)(x − b)):
- Assume the form A/(x − a) + B/(x − b).
- Multiply both sides by the denominator to remove fractions.
- Put x = a to find A, and put x = b to find B.
- Write the answer and check by adding back.
Case 2: repeated linear factors. A factor (x − a)² needs two terms:
- (x − a)² gives A/(x − a) + B/(x − a)².
- (x − a)³ gives A/(x − a) + B/(x − a)² + C/(x − a)³.
- Find the constants by substituting convenient values of x and by comparing the coefficients of like powers of x.
Case 3: an irreducible quadratic factor. A quadratic such as x² + 1 cannot be factorised over real numbers, and its numerator is linear:
- (ax² + bx + c) gives (Ax + B)/(ax² + bx + c).
- The other linear factors are treated as in Case 1.
- Find the constants by comparing coefficients, after using convenient values of x to find as many as possible.
The method works because two polynomials that are equal for all values of x must have equal coefficients of each power of x. This idea of equating coefficients is the basis of the work. The substitution method is a quick way to find a constant, because choosing x as the root of a factor makes the other terms vanish.
Always check the answer. Add the partial fractions back over a common denominator and see that the original numerator is obtained. A quick numerical check, putting x = 2 or x = 3 into both sides, can also expose a slip.
Study tip: Factorise the denominator completely before anything else, then decide which of the three cases applies. Use substitution for constants wherever you can, as it is faster, and use comparison of coefficients only for the rest. Always spend a minute on the check, because the exam question is usually worth full marks only when the final answer is exact.
Definitions and flashcards
Say the meaning first, then open the card to check.
Rational fraction
A fraction whose numerator and denominator are polynomials.
Proper fraction
A rational fraction with numerator degree less than denominator degree.
Improper fraction
A rational fraction with numerator degree equal to or greater than denominator degree.
Partial fractions
The simpler fractions whose sum is a given rational fraction.
Irreducible quadratic
A quadratic factor that cannot be factorised over the real numbers.
Identity
An equation true for every value of the variable.
Quotient
The polynomial obtained by dividing the numerator by the denominator.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Distinct linear factors | (px + q)/((x − a)(x − b)) = A/(x − a) + B/(x − b) | Case 1. |
| Repeated linear factor | N(x)/(x − a)² = A/(x − a) + B/(x − a)² | Case 2. |
| Irreducible quadratic | N(x)/((x − a)(x² + c)) = A/(x − a) + (Bx + C)/(x² + c) | Case 3. |
| Improper fraction | N(x)/D(x) = Q(x) + R(x)/D(x) | Divide first, then split the proper part. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
What is a proper rational fraction? Give one example. (2 marks)
A rational fraction is proper when the degree of the numerator is less than the degree of the denominator. For example, (x + 3)/(x² + 1) is proper because 1 < 2.
What is meant by resolving a fraction into partial fractions? (2 marks)
It means writing a single rational fraction as a sum of simpler fractions, each with a simpler denominator such as a linear factor of the original denominator.
What should be done first if the fraction is improper? (2 marks)
Divide the numerator by the denominator to get a quotient and a remainder. The remainder fraction is proper, and only that fraction is split into partial fractions.
Write the form of the partial fractions of 1/((x − 1)(x + 2)²). (3 marks)
There is one distinct linear factor (x − 1) and one repeated linear factor (x + 2)². The assumed form is A/(x − 1) + B/(x + 2) + C/(x + 2)². The constants A, B and C are then found by substitution and comparison of coefficients.
Why is the numerator over an irreducible quadratic factor taken as Bx + C and not just a constant? (3 marks)
The numerator must have degree one less than the denominator factor so that the fraction stays proper in general. A quadratic factor has degree 2, so the numerator may have degree 1, which is Bx + C. A constant alone would not give enough unknowns to match all the coefficients.
Resolve 5/((x − 1)(x + 4)) into partial fractions. (5 marks)
Let 5/((x − 1)(x + 4)) = A/(x − 1) + B/(x + 4). Multiply by the denominator: 5 = A(x + 4) + B(x − 1). Put x = 1: 5 = 5A, so A = 1. Put x = −4: 5 = −5B, so B = −1. Hence 5/((x − 1)(x + 4)) = 1/(x − 1) − 1/(x + 4). Check: (x + 4 − x + 1)/((x − 1)(x + 4)) = 5/((x − 1)(x + 4)).
Numericals with worked steps
Resolve (3x + 5)/((x + 1)(x + 2)) into partial fractions.
- Let (3x + 5)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2).
- 3x + 5 = A(x + 2) + B(x + 1).
- Put x = −1: 2 = A(1), so A = 2.
- Put x = −2: −1 = B(−1), so B = 1.
- Check: 2(x + 2) + (x + 1) = 3x + 5.
Answer: 2/(x + 1) + 1/(x + 2)
Resolve (2x + 3)/(x − 1)² into partial fractions.
- Let (2x + 3)/(x − 1)² = A/(x − 1) + B/(x − 1)².
- 2x + 3 = A(x − 1) + B.
- Put x = 1: 5 = B.
- Compare the coefficient of x: 2 = A.
- Check: 2(x − 1) + 5 = 2x + 3.
Answer: 2/(x − 1) + 5/(x − 1)²
Resolve 1/((x + 1)(x² + 1)) into partial fractions.
- Let 1/((x + 1)(x² + 1)) = A/(x + 1) + (Bx + C)/(x² + 1).
- 1 = A(x² + 1) + (Bx + C)(x + 1).
- Put x = −1: 1 = 2A, so A = 1/2.
- Compare x²: 0 = A + B, so B = −1/2.
- Compare constants: 1 = A + C, so C = 1/2.
- Check x: B + C = 0, which is true.
Answer: 1/(2(x + 1)) + (1 − x)/(2(x² + 1))
Resolve (x² + 1)/(x² − 1) into partial fractions.
- The fraction is improper. Divide: (x² + 1) = 1 × (x² − 1) + 2, so (x² + 1)/(x² − 1) = 1 + 2/(x² − 1).
- Let 2/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1).
- 2 = A(x + 1) + B(x − 1).
- Put x = 1: 2 = 2A, so A = 1.
- Put x = −1: 2 = −2B, so B = −1.
Answer: 1 + 1/(x − 1) − 1/(x + 1)
Common mistakes
- Trying to split an improper fraction without dividing first.
- Using only one term for a repeated factor. A factor (x − a)² needs both A/(x − a) and B/(x − a)².
- Writing a constant numerator over an irreducible quadratic instead of Bx + C.
- Making a sign error when substituting a root such as x = −2 and so getting the wrong constant.
- Not checking the answer by adding the fractions back.