This note shows how to split a single algebraic fraction into a sum of simpler fractions, called partial fractions. It is the reverse of adding fractions over a common denominator. Splitting a fraction this way makes later work, such as series expansions and integration, much easier.
Key ideas
A rational function is a fraction in which the top and bottom are polynomials. When the bottom (the denominator) factorises, the fraction can be written as a sum of fractions with simpler denominators. For example, \(\dfrac{1}{x-1}+\dfrac{2}{x+3}\) combines into one fraction with denominator \((x-1)(x+3)\). Partial fractions run this process backwards.
At this level the denominator has degree at most 3, and the fraction must be proper. That means the degree of the numerator is smaller than the degree of the denominator. The form you write down depends on the factors of the denominator.
- Distinct linear factors. Each factor \((x-a)\) gets its own constant: \(\dfrac{A}{x-a}+\dfrac{B}{x-b}\).
- A repeated linear factor. A factor \((x-a)^2\) needs two terms, one for each power: \(\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}\).
- An irreducible quadratic factor. A factor such as \(x^2+1\), which cannot be factorised over the real numbers, gets a linear numerator: \(\dfrac{Bx+C}{x^2+1}\).
Once the form is chosen, multiply both sides by the full denominator. This leaves an identity between polynomials that is true for every value of \(x\). There are two ways to find the unknown constants.
- Substitution. Choose values of \(x\) that make a factor zero. This removes some constants at once.
- Comparing coefficients. Expand the right-hand side and match the coefficients of \(x^2\), \(x\) and the constant term.
Often the best method is to combine both: substitute to get as many constants as you can, then compare coefficients for the rest.
If the fraction is improper, divide first so that you have a polynomial plus a proper fraction, then split the proper fraction.
You can always check an answer. Put the partial fractions back over a common denominator and see whether you recover the original numerator.
Method in four steps
Follow the same order each time. Having a fixed routine stops you from skipping the checks that catch careless mistakes, and it makes longer problems feel manageable. Work neatly, keeping the identity on its own line.
- Check that the fraction is proper, and factorise the denominator fully.
- Write the partial fraction form that matches each factor, using a new letter for each unknown.
- Multiply both sides by the original denominator to clear every fraction.
- Find the unknowns by substitution and by comparing coefficients, then write the final answer with the numbers in place.
Why does a repeated factor need two terms? With only \(\dfrac{B}{(x-2)^2}\), the numerator on the right would be a constant, but the original numerator can be a quadratic. Adding the term \(\dfrac{A}{x-2}\) gives enough freedom to match any numerator of lower degree than the denominator.
Here is how an improper fraction looks. For \(\dfrac{x^2+6x-1}{(x-1)(x+2)}\), the denominator expands to \(x^2+x-2\). Dividing gives a quotient of 1 and a remainder of \(5x+1\), so the fraction equals \(1+\dfrac{5x+1}{(x-1)(x+2)}\). The remaining fraction is the one from Example 1, so the full answer is \(1+\dfrac{2}{x-1}+\dfrac{3}{x+2}\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Proper fraction | The numerator has lower degree than the denominator. Only proper fractions can be split directly. |
| \(\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\) | Two distinct linear factors, with \(a\ne b\). |
| \(\dfrac{px^2+qx+r}{(x-a)(x-b)^2}=\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{(x-b)^2}\) | One distinct and one repeated linear factor. |
| \(\dfrac{px^2+qx+r}{(x-a)(x^2+c)}=\dfrac{A}{x-a}+\dfrac{Bx+C}{x^2+c}\) | A linear factor and an irreducible quadratic, with \(c\gt0\). |
| Identity | An equation that holds for all \(x\). This is why you may substitute any value of \(x\) once the denominators are cleared. |
Worked examples
Example 1 Express \(\dfrac{5x+1}{(x-1)(x+2)}\) in partial fractions.
Write \(\dfrac{5x+1}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}\).
Multiply through by \((x-1)(x+2)\): \(5x+1=A(x+2)+B(x-1)\).
Let \(x=1\): \(6=3A\), so \(A=2\).
Let \(x=-2\): \(-9=-3B\), so \(B=3\).
Answer: \(\dfrac{2}{x-1}+\dfrac{3}{x+2}\)
Example 2 Express \(\dfrac{5x^2-7x+6}{(x+1)(x-2)^2}\) in partial fractions.
The factor \((x-2)\) is repeated, so write \(\dfrac{A}{x+1}+\dfrac{B}{x-2}+\dfrac{C}{(x-2)^2}\).
Multiply through: \(5x^2-7x+6=A(x-2)^2+B(x+1)(x-2)+C(x+1)\).
Let \(x=-1\): \(5+7+6=9A\), so \(18=9A\) and \(A=2\).
Let \(x=2\): \(20-14+6=3C\), so \(12=3C\) and \(C=4\).
Compare coefficients of \(x^2\): \(5=A+B\), so \(B=5-2=3\).
Check with \(x=0\): the left side is 6. The right side is \(2(4)+3(1)(-2)+4(1)=8-6+4=6\). It matches.
Answer: \(\dfrac{2}{x+1}+\dfrac{3}{x-2}+\dfrac{4}{(x-2)^2}\)
Example 3 Express \(\dfrac{3x^2+5x+4}{(x+1)(x^2+1)}\) in partial fractions.
The factor \(x^2+1\) cannot be factorised, so use a linear numerator: \(\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+1}\).
Multiply through: \(3x^2+5x+4=A(x^2+1)+(Bx+C)(x+1)\).
Let \(x=-1\): \(3-5+4=2A\), so \(A=1\).
Compare \(x^2\): \(3=A+B\), so \(B=2\).
Compare constants: \(4=A+C\), so \(C=3\).
Check the \(x\) terms: the right side gives \(B+C=5\), which matches the 5 on the left.
Answer: \(\dfrac{1}{x+1}+\dfrac{2x+3}{x^2+1}\)
Common errors
- Writing only \(\dfrac{B}{(x-2)^2}\) for a repeated factor. You also need the term \(\dfrac{A}{x-2}\), otherwise the form cannot match every numerator.
- Using a constant numerator over an irreducible quadratic. It needs \(Bx+C\).
- Splitting an improper fraction directly. Divide first, or the constants will not exist.
- Substituting a value of \(x\) before clearing the denominators. Clear the fractions first, then substitute.
- Not checking. A quick substitution of an easy value, such as \(x=0\), catches most slips.
Exam tip: Use substitution first because it is fast, then compare coefficients only for the constants that remain. Always do a quick check with a spare value of \(x\).
Check yourself
- Express \(\dfrac{7}{(x-3)(x+4)}\) in partial fractions.
Answer
\(7=A(x+4)+B(x-3)\). At \(x=3\): \(7=7A\), so \(A=1\). At \(x=-4\): \(7=-7B\), so \(B=-1\). Result: \(\dfrac{1}{x-3}-\dfrac{1}{x+4}\). - Express \(\dfrac{3x+5}{(x+1)(x+2)}\) in partial fractions.
Answer
\(3x+5=A(x+2)+B(x+1)\). At \(x=-1\): \(2=A\). At \(x=-2\): \(-1=-B\), so \(B=1\). Result: \(\dfrac{2}{x+1}+\dfrac{1}{x+2}\). - Express \(\dfrac{3x+1}{(x-1)^2}\) in partial fractions.
Answer
Write \(\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}\), so \(3x+1=A(x-1)+B\). At \(x=1\): \(B=4\). Comparing \(x\): \(A=3\). Result: \(\dfrac{3}{x-1}+\dfrac{4}{(x-1)^2}\). - Express \(\dfrac{1}{x^2-1}\) in partial fractions.
Answer
\(x^2-1=(x-1)(x+1)\), so \(1=A(x+1)+B(x-1)\). At \(x=1\): \(A=\tfrac12\). At \(x=-1\): \(1=-2B\), so \(B=-\tfrac12\). Result: \(\dfrac{1}{2(x-1)}-\dfrac{1}{2(x+1)}\).