Algebra and functions

Q: What is the fundamental purpose of Partial Fraction Decomposition in P4?

It is the reverse process of adding algebraic fractions, used to break a single complex fraction into simpler parts that are easier to integrate.

Q: How do you identify if an algebraic fraction is "improper"?

An algebraic fraction is improper if the degree of the polynomial in the numerator is greater than or equal to the degree of the polynomial in the denominator.

Q: What must be done before applying partial fractions to an improper fraction?

You must perform algebraic (polynomial) division to express the fraction as a polynomial plus a proper fraction.

Q: What is the standard formula for representing the result of polynomial division?

$\frac{\text{Numerator}}{\text{Denominator}} = \text{Quotient} + \frac{\text{Remainder}}{\text{Denominator}}$

Q: What is the setup for a denominator with distinct linear factors?

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

Q: How is the setup for repeated linear factors different from distinct linear factors?

You must include a separate term for every power of the repeated factor, up to the highest power.

Q: What is the numerator setup for an irreducible quadratic factor?

The numerator must be a linear expression in the form $Ax + B$.

Q: What is the "Substitution Method" (Root Method) for finding constants?

A technique where you substitute specific values of $x$ (the roots of the factors) to eliminate terms and solve for unknown constants.

Q: When is the "Equating Coefficients" method necessary?

It is used when substitution cannot isolate all constants, specifically for repeated factors or quadratic factors.

Pure Mathematics (YPM01) · Pearson Edexcel International A Level · 7 min read · Updated: 6 Sept 2026

Welcome to Algebra and Functions (Unit P4)!

Hello mathematicians! This chapter is incredibly important. While you might feel like you've seen "algebra" and "functions" before, P4 takes these skills to the next level. We aren't just solving equations anymore—we are mastering algebraic manipulation so we can tackle the most advanced forms of calculus later in this unit, especially complex integration.

Don't worry if algebra sometimes feels like a puzzle; we will break down the trickiest P4 concept: Partial Fractions. By the end of this, you’ll be able to dismantle complicated fractions into smaller, simpler pieces, ready for integration!

The P4 Essential Tool: Partial Fractions

Partial Fraction Decomposition is the reverse process of adding algebraic fractions. You start with one complicated fraction and break it back down into the sum of two or more simpler fractions (its "parts").

Why Bother with Partial Fractions?

Imagine you have a difficult fraction to integrate, like $\int \frac{3x+1}{(x-1)(x+2)} dx$. Integrating this directly is very difficult.
However, if we can show that:
$$\frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$
...then the integration becomes simple, yielding terms involving ln (logarithms), which you’ll cover in the calculus section. Partial fractions turns a massive headache into two tiny, solvable problems.

Quick Review: Proper vs. Improper Fractions

Before you can apply partial fractions, you must check the degrees of the polynomials in the numerator (top) and the denominator (bottom).

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

Common Mistake to Avoid: Attempting to use partial fractions on an improper fraction without dividing first. This is a guaranteed way to lose marks!

Step-by-Step: Handling Improper Fractions (Polynomial Division)

If your fraction is improper, you must perform algebraic division to rewrite it as a polynomial plus a proper fraction.

Analogy: Think about the fraction $\frac{17}{3}$. You know $\frac{17}{3} = 5 + \frac{2}{3}$.

In algebra, we write the result of division like this: $$\frac{\text{Numerator}}{\text{Denominator}} = \text{Quotient} + \frac{\text{Remainder}}{\text{Denominator}}$$

How to Perform Algebraic Division

Let's divide $\frac{2x^3 + x^2 - 1}{x^2 + 1}$:

Setup: Write out the division (similar to long division with numbers).
First Term: Ask: What do I multiply $x^2$ by to get $2x^3$? Answer: $2x$. This is your first term of the Quotient.
Multiply Back: Multiply $2x$ by the divisor ($x^2 + 1$): $2x(x^2+1) = 2x^3 + 2x$.
Subtract: Subtract this result from the numerator. $(2x^3 + x^2 - 1) - (2x^3 + 2x) = x^2 - 2x - 1$.
Repeat: Ask: What do I multiply $x^2$ by to get the leading term of the remainder ($x^2$)? Answer: $+1$. This is the next term of the Quotient.
Final Subtraction: Multiply $1$ by the divisor: $1(x^2+1) = x^2 + 1$. Subtract this from the remainder: $(x^2 - 2x - 1) - (x^2 + 1) = -2x - 2$.
Result: Since the degree of $-2x-2$ (degree 1) is less than the degree of the divisor $x^2+1$ (degree 2), we stop.

The result is: $$\frac{2x^3 + x^2 - 1}{x^2 + 1} = \underbrace{(2x+1)}_{\text{Quotient}} + \underbrace{\frac{-2x-2}{x^2+1}}_{\text{Proper Fraction}}$$

You only apply partial fractions to the proper fraction part!

Key Takeaway: Always check the degrees. If improper, divide first. The partial fraction step only applies to the remainder term.

The Three Forms of Partial Fractions in P4

The structure of the denominator dictates how you set up the partial fraction decomposition. We use capital letters (A, B, C, etc.) as the unknown constants we need to find.

1. Case 1: Distinct Linear Factors

This is the most common and easiest type. The factors in the denominator are all different and linear (power 1).

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

Method for Finding A and B (Substitution Method - The Quick Trick)

The quickest way to solve for A and B is the substitution method (often called the 'cover-up' rule, though that name can be confusing).

Step 1: Set up the identity and multiply both sides by the original denominator: $$\frac{3x-1}{(x-2)(x+1)} \equiv \frac{A}{x-2} + \frac{B}{x+1}$$ $$3x-1 \equiv A(x+1) + B(x-2)$$

Step 2: Solve for A: Choose a value of x that makes the B term zero. (Let $x=2$). $$3(2)-1 = A(2+1) + B(2-2)$$ $$5 = 3A \implies A = \frac{5}{3}$$

Step 3: Solve for B: Choose a value of x that makes the A term zero. (Let $x=-1$). $$3(-1)-1 = A(-1+1) + B(-1-2)$$ $$-4 = -3B \implies B = \frac{4}{3}$$

Final Answer: $$\frac{3x-1}{(x-2)(x+1)} \equiv \frac{5/3}{x-2} + \frac{4/3}{x+1}$$

2. Case 2: Repeated Linear Factors

This occurs when one factor in the denominator is raised to a power greater than one (e.g., squared). This setup requires special attention.

Structure: You must include a term for every power of the repeated factor, up to the highest power. $$\frac{P(x)}{(ax+b)^2(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \frac{C}{cx+d}$$

Did you know? If the factor was cubed, $(ax+b)^3$, you would need three terms: $\frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \frac{C}{(ax+b)^3}$.

Method for Finding A, B, and C (Combination Method)

The substitution method only works easily for the factors whose roots clear the entire bracket (i.e., finding B and C). Finding A usually requires the Equating Coefficients method.

Consider: $$\frac{x^2+x+2}{(x-1)^2(x+1)} \equiv \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$$

Step 1: Multiply by the denominator, $(x-1)^2(x+1)$: $$x^2+x+2 \equiv A(x-1)(x+1) + B(x+1) + C(x-1)^2$$

Step 2: Use Substitution for B and C:

To find C, let $x = -1$: $$(-1)^2 + (-1) + 2 = C(-1-1)^2$$ $$2 = 4C \implies C = \frac{1}{2}$$
To find B, let $x = 1$: $$(1)^2 + (1) + 2 = B(1+1)$$ $$4 = 2B \implies B = 2$$

Step 3: Use Equating Coefficients for A:

Since substitution has failed to solve A easily, we look at the terms involving the highest power of $x$, which is $x^2$:

$$x^2+x+2 \equiv A(x^2-1) + B(x+1) + C(x^2-2x+1)$$

Equate the coefficients of $x^2$ on both sides:

LHS coefficient of $x^2$: 1
RHS coefficient of $x^2$: $A$ (from $A(x^2)$) + $C$ (from $C(x^2)$)

$$1 = A + C$$

Since we know $C = 1/2$: $$1 = A + \frac{1}{2} \implies A = \frac{1}{2}$$

The decomposition is: $$\frac{1/2}{x-1} + \frac{2}{(x-1)^2} + \frac{1/2}{x+1}$$

3. Case 3: Irreducible Quadratic Factors

While integration involving this case is usually outside the scope of P4, the decomposition itself is sometimes required. An Irreducible Quadratic Factor is one that cannot be factorised into real linear factors (i.e., $b^2 - 4ac < 0$). (e.g., $x^2+4$).

Structure: The numerator above an irreducible quadratic factor must be a linear expression ($Bx+C$).

$$\frac{P(x)}{(ax+b)(cx^2+dx+e)} \equiv \frac{A}{ax+b} + \frac{Bx+C}{cx^2+dx+e}$$

(Note: For P4 standard integration problems, you will primarily use Cases 1 and 2, but understanding this form is helpful for complex algebraic manipulation.)

Quick Review Box: Partial Fraction Setup Rules

1. Proper Check: Must divide first if improper.
2. Distinct Factors: $\frac{A}{ax+b} + \frac{B}{cx+d}$
3. Repeated Factors: $\frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \dots$ (Must include all powers up to the highest)
4. Irreducible Quadratic: $\frac{Ax+B}{x^2+c}$

Methods for Finding the Constants

You have two powerful techniques to find the constants (A, B, C...). Often, the best strategy is to use a combination of both!

Method A: Substitution (The Root Method)

This is the fastest method. By setting x equal to the root of a linear factor (e.g., $x=2$ for the factor $(x-2)$), you make all other terms containing that factor disappear, isolating one constant. This works perfectly for all distinct linear factors and the highest power term of a repeated linear factor.

Method B: Equating Coefficients

This is necessary when the substitution method cannot isolate all constants (e.g., for finding A in the repeated factor case, or whenever you deal with quadratic factors).

Process:

Expand the entire right-hand side of your identity.
Group terms based on powers of $x$ (e.g., $x^2$, $x^1$, $x^0$).
Match the coefficient of each power on the LHS to the corresponding total coefficient on the RHS.
Solve the resulting simultaneous equations.

Encouragement: Equating coefficients can be messy, but it is reliable. Always start by equating the coefficients of the highest power of $x$, as this often yields the simplest equation involving A, B, or C.

Summary and Next Steps

You have now mastered the key algebraic technique required for Pure Mathematics 4: Partial Fraction Decomposition. This skill is not abstract—it is a practical tool designed specifically to make integration possible.

Remember the P4 Mantra: Check degree, Divide if improper, Setup the factors correctly, Solve for constants. Good luck!

Quick check

Can you answer these now?

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

What is the fundamental purpose of Partial Fraction Decomposition in P4?

It is the reverse process of adding algebraic fractions, used to break a single complex fraction into simpler parts that are easier to integrate.

How do you identify if an algebraic fraction is "improper"?

An algebraic fraction is improper if the degree of the polynomial in the numerator is greater than or equal to the degree of the polynomial in the denominator.

What must be done before applying partial fractions to an improper fraction?

You must perform algebraic (polynomial) division to express the fraction as a polynomial plus a proper fraction.

What is the standard formula for representing the result of polynomial division?

$\frac{\text{Numerator}}{\text{Denominator}} = \text{Quotient} + \frac{\text{Remainder}}{\text{Denominator}}$

What is the setup for a denominator with distinct linear factors?

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

How is the setup for repeated linear factors different from distinct linear factors?

You must include a separate term for every power of the repeated factor, up to the highest power.

What is the numerator setup for an irreducible quadratic factor?

The numerator must be a linear expression in the form $Ax + B$.

What is the "Substitution Method" (Root Method) for finding constants?

A technique where you substitute specific values of $x$ (the roots of the factors) to eliminate terms and solve for unknown constants.

When is the "Equating Coefficients" method necessary?

It is used when substitution cannot isolate all constants, specifically for repeated factors or quadratic factors.

Ready to test yourself?

Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.

Practise This Topic

More Pure Mathematics (YPM01) chapters

* The content provided by thinka is generated by AI and may not always be accurate or up-to-date. Please use it as a supplementary resource and verify with official materials.

Put These Notes into Practice

Reading the notes is a great start. Now practise with unlimited AI-generated questions and get instant feedback. 100,000+ students are already improving their grades.

Start Practising Now View Pricing

Done reading? Test yourself with AI practice questions

Practise This Topic Now