Partial Fractions

students, have you ever seen a complicated fraction like $\frac{5x+1}{(x-2)(x+3)}$ and wondered how anyone could work with it? 😅 Partial fractions are a powerful algebra tool that break one complicated rational expression into simpler pieces. These simpler pieces are easier to add, integrate, compare, and analyze. In IB Mathematics: Analysis and Approaches HL, partial fractions help you move between factorised forms and expanded forms, and they connect directly to algebraic manipulation, functions, and calculus.

What partial fractions mean

A rational expression is a fraction of polynomials, such as $\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$. The goal of partial fractions is to rewrite a rational expression as a sum of simpler fractions.

For example,

$$\frac{5x+1}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}$$

for some constants $A$ and $B$. The expression on the right is easier to work with because each denominator is simpler.

Why does this matter? Because algebra often becomes easier when a complicated object is split into parts. This is especially useful when the denominator factors into linear terms, repeated factors, or irreducible quadratics. Partial fractions are part of the broader Number and Algebra topic because they rely on factorisation, symbolic manipulation, and solving equations. They also appear in calculus when integrating rational functions.

Key terminology

Proper rational expression: the degree of the numerator is less than the degree of the denominator.
Improper rational expression: the degree of the numerator is greater than or equal to the degree of the denominator.
Partial fraction decomposition: rewriting a rational expression as a sum of simpler fractions.
Linear factor: a factor of the form $x-a$.
Repeated factor: a factor appearing more than once, such as $(x-1)^2$.
Irreducible quadratic: a quadratic that cannot be factored over the real numbers, such as $x^2+1$.

How to decompose a rational expression

Before starting, always check whether the rational expression is proper. If it is improper, use polynomial division first. This is an important IB skill because it shows careful algebraic reasoning.

Example 1: distinct linear factors

Decompose

$$\frac{5x+1}{(x-2)(x+3)}$$

Step 1: Write the form of the decomposition.

$$\frac{5x+1}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}$$

Step 2: Multiply both sides by $(x-2)(x+3)$.

$$5x+1 = A(x+3) + B(x-2)$$

Step 3: Expand.

$$5x+1 = (A+B)x + (3A-2B)$$

Step 4: Match coefficients.

$$A+B=5$$

$$3A-2B=1$$

Solving gives $A=2$ and $B=3$. Therefore,

$$\frac{5x+1}{(x-2)(x+3)} = \frac{2}{x-2} + \frac{3}{x+3}$$

This result is the decomposition. It is easier to use than the original expression because each part has a simple denominator.

Example 2: repeated linear factor

Now consider

$$\frac{4x+7}{(x-1)^2(x+2)}$$

Because $(x-1)$ is repeated, the decomposition must include both powers:

$$\frac{4x+7}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$$

This structure is not random. For a repeated factor $(x-a)^n$, you must include fractions for every power from $1$ to $n$.

Multiply through by $(x-1)^2(x+2)$:

$$4x+7 = A(x-1)(x+2) + B(x+2) + C(x-1)^2$$

From here, you can expand and match coefficients, or use convenient values of $x$ to simplify. For example, $x=1$ and $x=-2$ make many terms disappear. This is a useful exam strategy because it reduces arithmetic errors 😊.

Working with improper rational expressions

If the numerator has degree at least as large as the denominator, you must first divide. This produces a polynomial plus a proper rational expression.

Example 3: polynomial division first

Decompose

$$\frac{x^2+3x+2}{x+1}$$

Since the numerator degree is larger than the denominator degree, divide first:

$$\frac{x^2+3x+2}{x+1} = x+2$$

because $(x+1)(x+2)=x^2+3x+2$.

In a less exact example, such as

$$\frac{x^2+1}{x-1},$$

you would write

$$\frac{x^2+1}{x-1} = x+1+\frac{2}{x-1}$$

Here the rational part is proper, so it can be further analyzed if needed. This step is essential in IB because it combines division, factorisation, and algebraic structure.

Irreducible quadratic factors

Sometimes the denominator contains a quadratic factor that cannot be broken into real linear factors. In that case, the numerator of the partial fraction must be linear.

For example,

$$\frac{2x+3}{(x-1)(x^2+4)}$$

has the form

$$\frac{2x+3}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4}$$

Why $Bx+C$ instead of just a constant? Because the numerator must be one degree less than the quadratic denominator. This rule keeps the decomposition flexible enough to match all possible numerators.

Example 4: quadratic factor decomposition

Multiply through by $(x-1)(x^2+4)$:

$$2x+3 = A(x^2+4) + (Bx+C)(x-1)$$

Expand the right side:

$$2x+3 = Ax^2+4A + Bx^2 - Bx + Cx - C$$

$$2x+3 = (A+B)x^2 + (C-B)x + (4A-C)$$

Match coefficients:

$$A+B=0$$

$$C-B=2$$

$$4A-C=3$$

Solving gives $A=\frac{1}{5}$, $B=-\frac{1}{5}$, and $C=\frac{9}{5}$. Therefore,

$$\frac{2x+3}{(x-1)(x^2+4)} = \frac{1/5}{x-1} + \frac{-\frac{1}{5}x+\frac{9}{5}}{x^2+4}$$

This decomposition is algebraically correct even though it looks more complicated at first. The advantage appears when you need to integrate or simplify further.

Why partial fractions are useful in IB Mathematics

Partial fractions connect strongly to several IB ideas:

Factorisation: you must factor the denominator carefully.
Solving equations: matching coefficients or substituting values creates systems of equations.
Functions: rational expressions are a key type of function.
Calculus: many integrals of rational functions become manageable after decomposition.
Proof and reasoning: the method depends on logical structure, not just memorised steps.

For example, when integrating

$$\int \frac{5x+1}{(x-2)(x+3)}\,dx,$$

using partial fractions gives

$$\int \left(\frac{2}{x-2}+\frac{3}{x+3}\right)\,dx = 2\ln|x-2|+3\ln|x+3|+C$$

This is much easier than trying to integrate the original fraction directly. This is one reason partial fractions are important in HL work.

Common mistakes and how to avoid them

students, here are some frequent errors students make:

Forgetting to check if the fraction is proper. If the degree of the numerator is too large, divide first.
Using the wrong decomposition form. Repeated linear factors need multiple terms; quadratic factors need linear numerators.
Expanding incorrectly. Small sign errors can change the final answer.
Not matching all coefficients. Make sure every power of $x$ is accounted for.
Ignoring domain restrictions. The original denominator cannot be zero.

A good habit is to check your answer by recombining the partial fractions. If they simplify back to the original expression, your decomposition is likely correct.

Conclusion

Partial fractions are a major algebra technique in IB Mathematics: Analysis and Approaches HL. They turn a complicated rational expression into a sum of simpler parts, making algebra, equation solving, and integration much easier. The method depends on factorisation, proper setup, and accurate coefficient matching. By understanding linear factors, repeated factors, and irreducible quadratics, students, you can use partial fractions confidently as part of the broader Number and Algebra toolkit. This topic is a clear example of how symbolic manipulation supports deeper mathematical reasoning and real problem solving ✨

Study Notes

A rational expression has the form $\frac{P(x)}{Q(x)}$ with $Q(x)\neq 0$.
If the rational expression is improper, use polynomial division first.
For distinct linear factors like $(x-a)(x-b)$, write fractions with constants in the numerators.
For repeated factors like $(x-a)^n$, include terms for every power from $1$ to $n$.
For an irreducible quadratic like $x^2+4$, use a linear numerator such as $Bx+C$.
Multiply through by the common denominator to create an identity.
Solve by matching coefficients or by substituting convenient values of $x$.
Check your final answer by recombining the fractions.
Partial fractions are useful in integration and in simplifying rational expressions.
This topic connects factorisation, algebraic structure, and functions in Number and Algebra.