3.1 Partial Fractions
AHL 1.11
- the difference between proper and improper algebraic fractions
- how to split a fraction like \(\dfrac{5x + 3}{x^2 + 2x - 3}\) into partial fractions
- two ways to find the unknown numerators: substituting values and equating coefficients
- (Extension) repeated factors, quadratic factors and improper fractions
The IB syllabus needs only denominators with two different linear factors, and a numerator of lower degree. The rest is marked Extension. You will use partial fractions again for integration (Unit 27) and for the binomial series (Unit 4).
Proper and improper fractions
An algebraic fraction \(\dfrac{f(x)}{g(x)}\) has polynomials on the top and the bottom.
Definition Not in the formula booklet
\(\dfrac{f(x)}{g(x)}\) is a proper fraction if the degree of \(f(x)\) is less than the degree of \(g(x)\). Otherwise it is an improper fraction.
The degree of a polynomial is its highest power of \(x\).
| Fraction | Degree of top | Degree of bottom | |
|---|---|---|---|
| \(\dfrac{4x - 9}{(x + 1)(x^2 + 3)}\) | 1 | 3 | proper |
| \(\dfrac{3x^2 + 1}{x^2 - 5}\) | 2 | 2 | improper |
| \(\dfrac{x^4}{(x - 2)(x + 7)}\) | 4 | 2 | improper |
Splitting a fraction
You already know how to add fractions:
\[ \frac{3}{x + 3} + \frac{2}{x - 1} = \frac{3(x - 1) + 2(x + 3)}{(x + 3)(x - 1)} = \frac{5x + 3}{(x + 3)(x - 1)} . \]
Partial fractions do the opposite: we start with \(\dfrac{5x + 3}{(x + 3)(x - 1)}\) and split it back into \(\dfrac{3}{x + 3} + \dfrac{2}{x - 1}\).
Two distinct linear factors (in the syllabus) Not in the formula booklet
\[ \frac{px + q}{(ax + b)(cx + d)} \equiv \frac{A}{ax + b} + \frac{B}{cx + d} \]
- \(p\): a constant (the coefficient of \(x\) in the numerator)
- \(q\): a constant (the constant term of the numerator)
- \(a\): a constant (the coefficient of \(x\) in the first linear factor)
- \(b\): a constant (the constant term of the first linear factor)
- \(c\): a constant (the coefficient of \(x\) in the second linear factor)
- \(d\): a constant (the constant term of the second linear factor)
- \(A\): a constant to be found
- \(B\): a constant to be found
The sign \(\equiv\) means the two sides are equal for every value of \(x\) (an identity, see Unit 6). That is why we may substitute any value we like.
Method
- Factorise the denominator.
- Write the partial fractions with unknown numerators \(A\), \(B\).
- Multiply both sides by the whole denominator.
- Find \(A\) and \(B\) by substituting values of \(x\) that make a bracket zero, or by equating coefficients of \(x\) and the constant terms.
- Check by adding the partial fractions back together (or by testing one value of \(x\)).
Example 1 Express \(\dfrac{5x + 3}{x^2 + 2x - 3}\) in partial fractions.
Factorise: \(x^2 + 2x - 3 = (x + 3)(x - 1)\). Write
\[ \frac{5x + 3}{(x + 3)(x - 1)} \equiv \frac{A}{x + 3} + \frac{B}{x - 1} . \]
Multiply both sides by \((x + 3)(x - 1)\):
\[ 5x + 3 \equiv A(x - 1) + B(x + 3) . \]
Put \(x = 1\) (this removes \(A\)): \(\;8 = 4B\), so \(B = 2\).
Put \(x = -3\) (this removes \(B\)): \(\;-12 = -4A\), so \(A = 3\).
\[ \frac{5x + 3}{x^2 + 2x - 3} = \frac{3}{x + 3} + \frac{2}{x - 1} \]
Example 2 Express \(\dfrac{8x - 1}{(2x - 1)(x + 1)}\) in partial fractions. Find the constants in two different ways.
\[ \frac{8x - 1}{(2x - 1)(x + 1)} \equiv \frac{A}{2x - 1} + \frac{B}{x + 1} \quad\Longrightarrow\quad 8x - 1 \equiv A(x + 1) + B(2x - 1) . \]
Method 1 — substitution. \(x = \frac{1}{2}\): \(\;3 = \frac{3}{2}A\), so \(A = 2\). \(\quad\) \(x = -1\): \(\;-9 = -3B\), so \(B = 3\).
Method 2 — equating coefficients. Expand the right-hand side: \((A + 2B)x + (A - B)\).
\[ \text{coefficients of } x:\; A + 2B = 8, \qquad \text{constants}:\; A - B = -1 . \]
Subtracting: \(3B = 9\), so \(B = 3\) and \(A = 2\). Both methods give
\[ \frac{8x - 1}{(2x - 1)(x + 1)} = \frac{2}{2x - 1} + \frac{3}{x + 1} . \]
Substitution is usually quicker. Equating coefficients is useful when no value of \(x\) makes things simple.
Extension — other kinds of denominator
These are not required by the IB syllabus, but they appear in some integration problems.
| Factor in the denominator | Partial fractions to use |
|---|---|
| distinct linear: \((ax + b)(cx + d)\) | \(\dfrac{A}{ax + b} + \dfrac{B}{cx + d}\) |
| repeated linear: \((ax + b)^2\) | \(\dfrac{A}{ax + b} + \dfrac{B}{(ax + b)^2}\) |
| quadratic that cannot be factorised: \(x^2 + c^2\) | \(\dfrac{Ax + B}{x^2 + c^2}\) |
For an improper fraction, first divide, to write it as a polynomial plus a proper fraction.
Example A — repeated factor. Express \(\dfrac{x^2 + 7x}{(x - 1)(x + 1)^2}\) in partial fractions.
\[ x^2 + 7x \equiv A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1) \]
\(x = 1\): \(8 = 4A\), \(A = 2\). \(\;x = -1\): \(-6 = -2C\), \(C = 3\). Coefficients of \(x^2\): \(1 = A + B\), so \(B = -1\).
\[ \frac{x^2 + 7x}{(x - 1)(x + 1)^2} = \frac{2}{x - 1} - \frac{1}{x + 1} + \frac{3}{(x + 1)^2} \]
Example B — quadratic factor. Express \(\dfrac{4x^2 - 1}{(x - 2)(x^2 + 1)}\) in partial fractions.
\[ 4x^2 - 1 \equiv (Ax + B)(x - 2) + C(x^2 + 1) \]
\(x = 2\): \(15 = 5C\), \(C = 3\). \(\;x = 0\): \(-1 = -2B + 3\), \(B = 2\). Coefficients of \(x^2\): \(4 = A + C\), so \(A = 1\).
\[ \frac{4x^2 - 1}{(x - 2)(x^2 + 1)} = \frac{x + 2}{x^2 + 1} + \frac{3}{x - 2} \]
Example C — improper fraction. Express \(\dfrac{2x^2 + 5x - 1}{(x + 2)(x - 1)}\) in partial fractions.
The denominator is \(x^2 + x - 2\). Divide: \(2x^2 + 5x - 1 = 2(x^2 + x - 2) + (3x + 3)\), so
\[ \frac{2x^2 + 5x - 1}{(x + 2)(x - 1)} = 2 + \frac{3x + 3}{(x + 2)(x - 1)} = 2 + \frac{1}{x + 2} + \frac{2}{x - 1} . \]
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Not factorising the denominator first | \(x^2 + 2x - 3 = (x + 3)(x - 1)\) | The partial fractions come from the factors. |
| Splitting an improper fraction directly | Divide first | Partial fractions of the form \(\frac{A}{\ldots} + \frac{B}{\ldots}\) only give proper fractions. |
| \(x = \frac{1}{2}\) gives \(A = \frac{3}{2} \times 3\) | \(3 = \frac{3}{2}A \Rightarrow A = 2\) | Divide, do not multiply, to find \(A\). |
| Forgetting to check | Add back, or test \(x = 0\) | A quick check finds most mistakes. |
Exercises
1 Express \(\dfrac{5}{(x + 1)(x + 6)}\) in partial fractions.
\(5 \equiv A(x + 6) + B(x + 1)\). \(x = -1\): \(A = 1\). \(x = -6\): \(B = -1\).
\[ \frac{1}{x + 1} - \frac{1}{x + 6} \]
2 Express \(\dfrac{x + 7}{(x - 1)(x + 3)}\) in partial fractions.
\(x + 7 \equiv A(x + 3) + B(x - 1)\). \(x = 1\): \(8 = 4A\), \(A = 2\). \(x = -3\): \(4 = -4B\), \(B = -1\).
\[ \frac{2}{x - 1} - \frac{1}{x + 3} \]
3 Express \(\dfrac{4x}{x^2 - 4}\) in partial fractions.
\(x^2 - 4 = (x - 2)(x + 2)\). \(4x \equiv A(x + 2) + B(x - 2)\): \(x = 2\) gives \(A = 2\), \(x = -2\) gives \(B = 2\).
\[ \frac{2}{x - 2} + \frac{2}{x + 2} \]
4 Express \(\dfrac{7x + 1}{2x^2 + x - 1}\) in partial fractions.
\(2x^2 + x - 1 = (2x - 1)(x + 1)\). \(7x + 1 \equiv A(x + 1) + B(2x - 1)\).
\(x = \frac{1}{2}\): \(\frac{9}{2} = \frac{3}{2}A\), \(A = 3\). \(\;x = -1\): \(-6 = -3B\), \(B = 2\).
\[ \frac{3}{2x - 1} + \frac{2}{x + 1} \]
5 Express \(\dfrac{3x - 4}{x^2 - x - 6}\) in partial fractions.
\(x^2 - x - 6 = (x - 3)(x + 2)\). \(3x - 4 \equiv A(x + 2) + B(x - 3)\).
\(x = 3\): \(5 = 5A\), \(A = 1\). \(x = -2\): \(-10 = -5B\), \(B = 2\).
\[ \frac{1}{x - 3} + \frac{2}{x + 2} \]
6 Express \(\dfrac{1}{x(x + 1)}\) in partial fractions.
\(1 \equiv A(x + 1) + Bx\): \(x = 0\) gives \(A = 1\); \(x = -1\) gives \(B = -1\). \(\;\dfrac{1}{x} - \dfrac{1}{x + 1}\).
7 Given that \(\dfrac{ax + b}{(x + 1)(x + 2)} \equiv \dfrac{3}{x + 1} - \dfrac{1}{x + 2}\), find \(a\) and \(b\).
Add the right-hand side: \(\dfrac{3(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \dfrac{2x + 5}{(x + 1)(x + 2)}\). So \(a = 2\), \(b = 5\).
8 Express \(\dfrac{5x + 1}{(3x - 1)(x + 1)}\) in partial fractions.
\(5x + 1 \equiv A(x + 1) + B(3x - 1)\).
\(x = \frac{1}{3}\): \(\frac{8}{3} = \frac{4}{3}A\), \(A = 2\). \(\;x = -1\): \(-4 = -4B\), \(B = 1\).
\[ \frac{2}{3x - 1} + \frac{1}{x + 1} \]
9 Extension Express \(\dfrac{3x + 1}{(x + 1)^2}\) in partial fractions.
\(3x + 1 \equiv A(x + 1) + B\). Coefficient of \(x\): \(A = 3\). \(x = -1\): \(B = -2\).
\[ \frac{3}{x + 1} - \frac{2}{(x + 1)^2} \]
10 Extension Express \(\dfrac{x^2 + 3}{(x - 1)(x + 1)}\) in partial fractions.
Improper: \(x^2 + 3 = (x^2 - 1) + 4\), so the fraction is \(1 + \dfrac{4}{(x - 1)(x + 1)}\).
\(4 \equiv A(x + 1) + B(x - 1)\): \(A = 2\), \(B = -2\).
\[ 1 + \frac{2}{x - 1} - \frac{2}{x + 1} \]