Algebra and functions

Q: How do you distinguish between a proper and an improper algebraic fraction?

A fraction is proper if the degree of the numerator is strictly less than the degree of the denominator. It is improper if the degree of the numerator is greater than or equal to the degree of the denominator.

Q: What must you do if you encounter an improper algebraic fraction?

You must perform algebraic long division to rewrite the expression as a polynomial (the quotient) plus a proper fraction.

Mathematics (YMA01) · Pearson Edexcel International A Level · 7 min read · Updated: 7 Sept 2026

Welcome to Algebra and Functions: Partial Fractions!

Hello future mathematician! This chapter, Algebra and Functions, might seem small in P4, but it contains one of the most powerful algebraic tools you will learn: Partial Fractions.

Don't worry if fractions feel intimidating—we are going to break them down! (Literally.) This technique is crucial because it allows us to take complicated fractions that are impossible to integrate directly (a key skill you need in P4) and turn them into simpler, manageable pieces. Think of it as breaking a difficult puzzle into smaller, solvable mini-puzzles.

Let's master the art of algebraic decomposition!

1. Understanding Partial Fraction Decomposition

1.1 What are Partial Fractions?

When you add two simple fractions together, you get a single, more complicated fraction. Partial Fraction Decomposition is simply the reverse process. We take a single complicated algebraic fraction and split it back up into the simpler fractions that created it.

Key Concept: We only use partial fractions when the denominator of the algebraic fraction can be factorised into linear factors (e.g., $ (x-2) $ or $ (3x+1) $).

Why do we need this in P4?

The primary use of partial fractions in Pure Mathematics 4 is to prepare complex rational functions for integration. Integrating $ \frac{1}{(x+1)(x-2)} $ is very difficult, but integrating $ \frac{A}{x+1} + \frac{B}{x-2} $ becomes straightforward using the logarithmic integral rule (which you learned in P3).

2. The Crucial Prerequisite: Checking the Degree

Before you even start decomposing, you must check the relationship between the degree (the highest power of $x$) of the numerator (top) and the denominator (bottom).

2.1 Proper vs. Improper Fractions

Proper Fraction: The degree of the numerator is less than the degree of the denominator. (e.g., $ \frac{x+1}{x^2+3x} $). You can proceed straight to partial fraction decomposition.
Improper Fraction: The degree of the numerator is greater than or equal to the degree of the denominator. (e.g., $ \frac{x^2+5}{x^2-1} $ or $ \frac{x^3}{x^2+x} $). You MUST perform algebraic division first!

2.2 Algebraic Long Division (Dealing with Improper Fractions)

If you have an improper fraction, you must divide the numerator by the denominator. This process is very similar to numerical long division.

The Goal: To rewrite the improper fraction in the form:
$$ \text{Original Fraction} = \text{Quotient} + \frac{\text{Remainder}}{\text{Original Denominator}} $$

The quotient is usually a simple polynomial (like $ A $ or $ Ax+B $), and the remaining fraction will now be a proper fraction that you can decompose using partial fractions.

Step-by-Step Example (Conceptual):
Let's simplify $ \frac{x^2+3x+5}{x+1} $.

Divide the leading terms: $ x^2 \div x = x $. (This is the first term of the Quotient.)
Multiply the quotient term by the divisor: $ x(x+1) = x^2+x $.
Subtract this result from the numerator: $ (x^2+3x+5) - (x^2+x) = 2x+5 $.
Repeat: Divide the new leading term $ 2x $ by $ x $: $ 2x \div x = 2 $. (This is the second term of the Quotient.)
Multiply and subtract: $ (2x+5) - 2(x+1) = (2x+5) - (2x+2) = 3 $. (This is the Remainder.)

Result: $ \frac{x^2+3x+5}{x+1} = x + 2 + \frac{3}{x+1} $.

Quick Review: Improper Fractions

Check: If Num Degree ≥ Denom Degree, Divide first!

Avoid the Mistake: Never attempt partial fractions on an improper fraction before division. The decomposition will fail.

3. Decomposition Techniques: The Two Essential Cases

Once you have a proper fraction, you can start the decomposition. We use capital letters (A, B, C, etc.) to represent the unknown constants we need to find.

3.1 Case 1: Distinct Linear Factors

This is the simplest and most common case. The denominator is composed of different, non-repeated linear factors.

Form: $$ \frac{P(x)}{(ax+b)(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{cx+d} $$

Step-by-Step Process (Example: $ \frac{5x-1}{(x-1)(x+2)} $):

Set up the identity: $$ \frac{5x-1}{(x-1)(x+2)} \equiv \frac{A}{x-1} + \frac{B}{x+2} $$
Combine the RHS (Right-Hand Side): Multiply by the common denominator $(x-1)(x+2)$. $$ 5x-1 \equiv A(x+2) + B(x-1) $$
Find A and B using the Substitution Method: This is the quickest way! Choose values of $x$ that make the factors equal to zero.
- To find A, let $ x = 1 $ (This makes the $B$ term disappear): $$ 5(1) - 1 = A(1+2) + B(1-1) $$ $$ 4 = 3A \implies A = \frac{4}{3} $$
- To find B, let $ x = -2 $ (This makes the $A$ term disappear): $$ 5(-2) - 1 = A(-2+2) + B(-2-1) $$ $$ -11 = -3B \implies B = \frac{11}{3} $$
Write the final answer: $$ \frac{5x-1}{(x-1)(x+2)} = \frac{4/3}{x-1} + \frac{11/3}{x+2} $$

Memory Aid (The Substitution Trick): Always pick the $x$ value that "kills" (turns to zero) the term you don't want to calculate yet. This is much faster than equating coefficients for distinct factors.

3.2 Case 2: Repeated Linear Factors

This case involves a linear factor raised to a power (usually squared, but sometimes cubed). This requires an extra constant. Students often miss this factor!

If you have $ (ax+b)^2 $ in the denominator, you MUST include both $ (ax+b) $ and $ (ax+b)^2 $ in your decomposition.

Form: $$ \frac{P(x)}{(ax+b)^2(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \frac{C}{cx+d} $$

Strategy: Use substitution for the distinct factor and the highest power repeated factor. Then use the Equating Coefficients Method for the remaining constant(s).

Step-by-Step Process (Example: $ \frac{x^2+1}{x(x+1)^2} $):

Set up the identity: $$ \frac{x^2+1}{x(x+1)^2} \equiv \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} $$
Clear Denominators: Multiply by $ x(x+1)^2 $. $$ x^2+1 \equiv A(x+1)^2 + Bx(x+1) + Cx $$
Use Substitution (Find A and C easily):
- To find A, let $ x = 0 $: $$ (0)^2+1 = A(0+1)^2 + B(0) + C(0) $$ $$ 1 = A(1) \implies \mathbf{A = 1} $$
- To find C, let $ x = -1 $: $$ (-1)^2+1 = A(0) + B(0) + C(-1) $$ $$ 2 = -C \implies \mathbf{C = -2} $$
Use Equating Coefficients (Find B): Since we have used the substitution values, we must now equate coefficients. The easiest way is usually to look at the highest power, $ x^2 $.
Expand the RHS of the identity: $$ x^2+1 \equiv A(x^2+2x+1) + B(x^2+x) + Cx $$ Equate the coefficients of $ x^2 $: $$ \text{LHS coefficient: } 1 $$ $$ \text{RHS coefficients: } A + B $$ $$ 1 = A + B $$ Since we know $ A=1 $: $$ 1 = 1 + B \implies \mathbf{B = 0} $$
Write the final answer: $$ \frac{x^2+1}{x(x+1)^2} = \frac{1}{x} + \frac{0}{x+1} + \frac{-2}{(x+1)^2} = \frac{1}{x} - \frac{2}{(x+1)^2} $$

⚠️ Common Mistake Alert (Repeated Factors)

Struggling students often forget to include the factor to the power of one.
Incorrect Setup: $ \frac{A}{x} + \frac{B}{(x+1)^2} $
Correct Setup: $ \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} $
Remember: You must account for all possible denominators that could have led to the final fraction.

4. The Link to Integration (P4 Application)

Did you know? Partial fractions is often called a "pre-integration" technique in P4.

Once a fraction has been decomposed, its integration becomes trivial, relying heavily on two key results:

Logarithmic Rule (for linear denominators): $$ \int \frac{k}{ax+b} \, dx = \frac{k}{a} \ln|ax+b| + C $$
Reverse Chain Rule (for repeated factors): $$ \int \frac{k}{(ax+b)^n} \, dx = k \int (ax+b)^{-n} \, dx $$ We treat this as a standard polynomial integration where $ u = ax+b $.

Example of Integration:
If we integrated the result from Case 1: $ \frac{4/3}{x-1} + \frac{11/3}{x+2} $ $$ \int \left( \frac{4/3}{x-1} + \frac{11/3}{x+2} \right) dx = \frac{4}{3} \ln|x-1| + \frac{11}{3} \ln|x+2| + C $$

Key Takeaways and Final Review

You have now mastered the decomposition of algebraic fractions! Here are the core steps to remember for any problem in this chapter:

The Partial Fractions Checklist

Check the Degree: Is the fraction improper ($ \text{deg}(N) \geq \text{deg}(D) $)? If yes, perform Algebraic Long Division first.
Factorise: Fully factorise the denominator into its linear factors.
Set up the Identity:
- Distinct Factors: Use simple $ \frac{A}{\text{factor}} $ form.
- Repeated Factors (e.g., $ (x-a)^2 $): Use both $ \frac{A}{x-a} $ AND $ \frac{B}{(x-a)^2} $.
Solve for Constants: Use the Substitution Method for quick calculations (setting $x$ to eliminate terms), and the Equating Coefficients Method for any remaining constants (especially necessary for repeated factors).
Final Form: Write the fraction as the sum of the partial fractions (plus any quotient from step 1, if it was improper).

Don't worry if equating coefficients takes time; just be methodical and careful with your algebra!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is Partial Fraction Decomposition?

It is the process of splitting a single complex algebraic fraction into the sum of two or more simpler fractions, essentially reversing the process of adding fractions with a common denominator.

What condition must the denominator meet to be decomposed into partial fractions?

The denominator of the algebraic fraction must be factorisable into linear factors, such as $ (ax + b) $.

How do you distinguish between a proper and an improper algebraic fraction?

A fraction is proper if the degree of the numerator is strictly less than the degree of the denominator. It is improper if the degree of the numerator is greater than or equal to the degree of the denominator.

What must you do if you encounter an improper algebraic fraction?

You must perform algebraic long division to rewrite the expression as a polynomial (the quotient) plus a proper fraction.

What is the identity form for a fraction with distinct linear factors?

For distinct factors in the denominator, the form is: $ \frac{P(x)}{(ax+b)(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{cx+d} $.

How does the substitution method help solve for constants in partial fractions?

By substituting specific values of $ x $ that make individual linear factors equal to zero, you can eliminate all terms except the one containing the constant you wish to find.

What is the identity form for a fraction with repeated linear factors?

If a factor is repeated (e.g., squared), you must include a term for every power of that factor: $ \frac{A}{ax+b} + \frac{B}{(ax+b)^2} $.

When and how do you use the 'Equating Coefficients' method?

It is used when substitution alone isn't enough; you expand the right-hand side and match the coefficients of $ x^2 $, $ x $, or the constant terms from both sides of the identity.

What is the standard integral for a partial fraction with a linear denominator?

The integral follows the logarithmic rule: $ \int \frac{k}{ax+b} \, dx = \frac{k}{a} \ln|ax+b| + C $.

How do you integrate a partial fraction with a squared linear denominator?

You rewrite it using a negative index and use the reverse chain rule: $ \int k(ax+b)^{-2} \, dx = \frac{k}{a(-1)}(ax+b)^{-1} + C $.

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