Maths › Algebra and functions › Partial fractions
Partial fractions
Adding fractions is easy and everyone learns it forwards. This lesson runs it backwards, splitting one awkward algebraic fraction into simple pieces with unknown constants on top, found by substituting cunning values of x. The payoff arrives later, when integration and series expansions want the pieces.
Builds on Polynomials and the factor theorem.
IN THIS TOPIC
- Split a fraction whose denominator is two or three distinct linear brackets.
- Handle a repeated linear factor with the two-fraction template.
COMMON MISCONCEPTION
A squared bracket in the denominator needs just one fraction.
Un-adding fractions
Add 2/(x + 1) and 3/(x + 2) over a common denominator and you get (5x + 7)/((x + 1)(x + 2)). Partial fractions is that computation reversed. Write the split with unknown constants, multiply up, and choose values of x that silence one bracket at a time.
WORKED EXAMPLE
Two brackets, two constants
Express (5x + 7)/((x + 1)(x + 2)) in partial fractions.
Set (5x + 7)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2) and multiply up to get 5x + 7 = A(x + 2) + B(x + 1).
Substitute x = −1 to eliminate the B bracket, giving 2 = A. Substitute x = −2 to eliminate the A bracket, giving −3 = −B, so B = 3.
So the fraction is 2/(x + 1) + 3/(x + 2).
Any spare value of x in the common domain, avoiding the excluded −1 and −2, checks the answer. At x = 0 the left side is 7/2 and the right side is 2 + 3/2, which agree.
GUIDED PRACTICE
Three brackets, three constants
Express (6x2 + 5x − 2)/(x(x − 1)(2x + 1)) in partial fractions, before opening the working.
Show the working
The template is A/x + B/(x − 1) + C/(2x + 1), and multiplying up gives 6x2 + 5x − 2 = A(x − 1)(2x + 1) + Bx(2x + 1) + Cx(x − 1).
x = 0 gives −2 = −A, so A = 2. x = 1 gives 9 = 3B, so B = 3. x = −½ gives −3 = ¾C, so C = −4.
The fraction is 2/x + 3/(x − 1) − 4/(2x + 1).
Each substitution was a bracket's own root. That is what silenced two of the three terms at a stroke.
The repeated factor
A squared bracket breaks the one-fraction-per-bracket pattern. The denominator (cx + d)2 can shelter two distinct simple pieces, one over the bracket and one over its square, and in general both are needed.
INDEPENDENT PRACTICE
A squared bracket, in full
Express (x + 4)/((x + 1)(x − 2)2) in partial fractions.
Show the working
Template: A/(x + 1) + B/(x − 2) + C/(x − 2)2, so x + 4 = A(x − 2)2 + B(x + 1)(x − 2) + C(x + 1).
x = −1 gives 3 = 9A, so A = ⅓. x = 2 gives 6 = 3C, so C = 2.
No value of x eliminates B alone, so compare a coefficient instead. The constant terms give 4 = 4A − 2B + C, and with A and C known, B = −⅓.
The fraction is ⅓/(x + 1) − ⅓/(x − 2) + 2/(x − 2)2. Substituting x = 1 into the multiplied-up identity gives 5 on both sides, which confirms it.
Substitution reaches the constants sitting at the brackets' roots. The leftover constant always needs a compared coefficient or one extra substitution.
ASSESSMENT FOCUS
- Write the template first. One fraction per distinct bracket, two for a squared one. That line carries a mark of its own, ahead of all the arithmetic.
- Multiply up and substitute each bracket's root, which picks the constants off one at a time. When a constant survives every substitution, compare coefficients; the constant term is usually quickest.
- Keep the constants as exact fractions, because thirds and quarters are normal here and decimals are marked down. A spare value of x checks the lot in ten seconds.
CHECK YOURSELF
Express (7x − 1)/((x − 1)(x + 3)) in partial fractions.
Show a hint
Two brackets, two constants, two substitutions.
Show the answer
Template and multiply up. 7x − 1 = A(x + 3) + B(x − 1).
x = 1 gives 6 = 4A, so A = 3/2. x = −3 gives −22 = −4B, so B = 11/2.
The fraction is 3/(2(x − 1)) + 11/(2(x + 3)), and x = 0 checks it, since 1/3 = −3/2 + 11/6.
One fraction per distinct bracket, and a squared bracket takes both the bracket and its square.
Substitute the roots to harvest constants, then compare a coefficient for whatever is left.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the partial fractions questions page.
CHECK YOUR PROGRESS
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- Split a fraction whose denominator is two or three distinct linear brackets.
- Handle a repeated linear factor with the two-fraction template.
Open the full revision checklist to see every objective in the course in one place.