# Algebra

> CIE A-Level Mathematics · Pure Mathematics 2 & 3
> Source: https://www.owlsprep.com/study/cie-9709-u2-algebra/

This module covers the core algebra of Pure 2 and 3: the modulus function, polynomial division, and the factor and remainder theorems (shared by Papers 2 and 3), then the Paper 3 topics of partial fractions and the binomial expansion of rational functions with negative and fractional powers.

**Prerequisites:** [CIE A-Level Pure 1 Quadratics](https://www.owlsprep.com/study/cie-9709-u1-quadratics/); [Binomial expansion for positive integer powers](https://www.owlsprep.com/study/cie-9709-u1-series/)

## Learning objectives

- Interpret and sketch the modulus function y = |ax + b|, and solve modulus equations and inequalities
- Divide a polynomial (degree up to 4) by a linear or quadratic divisor, identifying the quotient and remainder
- Use the factor and remainder theorems to factorise polynomials and find unknown coefficients
- Distinguish between proper and improper rational functions and simplify improper fractions
- Decompose proper and improper rational functions into partial fractions
- Apply the binomial expansion to negative and fractional powers with correct validity

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P2 past papers (2016–2025): **Algebra** appears **250 times** in the last 10 years — **27.5%** of all P2 questions (250 of 455).

Most-tested forms: Polynomial division and factor/remainder theorems (93), Modulus function: |ax+b| equations and inequalities (89), Partial fractions (linear, repeated linear, quadratic) (53), Binomial expansion for any rational n; range of validity (40).

## The Modulus Function

**Modulus (Absolute Value)** — The modulus $|x|$ is the magnitude of $x$: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$. It is never negative and measures the distance of $x$ from $0$ on the number line.

*Notation:* |x|

*Example:* $|5| = 5$, $|-5| = 5$, and $|x - 3|$ is the distance between $x$ and $3$

The graph of $y = |ax + b|$ is a **V-shape**. Draw the line $y = ax + b$, then reflect the part that lies **below** the $x$-axis up above it. The corner (vertex) sits on the $x$-axis where $ax + b = 0$, i.e. at $x = -\frac{b}{a}$, and the two arms have gradients $a$ and $-a$.

**Key Modulus Relations** — Two useful equivalences: $|a| = |b| \iff a^2 = b^2$ (true for all $a$ and $b$), which lets you remove modulus signs by squaring; and, for $b > 0$, $|x - a| < b \iff a - b < x < a + b$, which turns a modulus inequality into a double inequality.

*Notation:* |a| = |b| \iff a^2 = b^2

*Example:* $|x - 4| < 3 \iff 1 < x < 7$

**Worked example:** Solve $|3x - 2| = |2x + 7|$.

1. Both sides are moduli, so square both sides to remove them (using $|a| = |b| \iff a^2 = b^2$):
2. $$(3x - 2)^2 = (2x + 7)^2$$
3. Expand each side:
4. $$9x^2 - 12x + 4 = 4x^2 + 28x + 49$$
5. Bring everything to one side and simplify:
6. $$5x^2 - 40x - 45 = 0 \implies x^2 - 8x - 9 = 0$$
7. Factorise and solve:
8. $$(x - 9)(x + 1) = 0 \implies x = 9 \ \text{or} \ x = -1$$

**Worked example:** Solve the inequality $|2x - 1| < 5$.

1. A single modulus less than a positive number becomes a double inequality, $-5 < 2x - 1 < 5$:
2. $$-5 < 2x - 1 < 5$$
3. Add $1$ throughout, then divide by $2$:
4. $$-4 < 2x < 6 \implies -2 < x < 3$$

> **Exam tip:** With a modulus on BOTH sides, square both sides; with a modulus on only one side, either square (when both sides are non-negative) or split into the two cases $\pm$.

## Polynomial Division

**Division Identity** — Dividing a polynomial $P(x)$ (degree up to 4) by a divisor $D(x)$ gives a **quotient** $Q(x)$ and a **remainder** $R(x)$ with $\deg R < \deg D$. If $R(x) = 0$ then $D(x)$ is a factor of $P(x)$. Dividing by a linear divisor leaves a constant remainder; dividing by a quadratic leaves a remainder of the form $px + q$.

*Notation:* P(x) = D(x)\,Q(x) + R(x)

*Example:* $x^2 - 5x + 6 = (x - 2)(x - 3) + 0$, so $(x - 2)$ is a factor

You can divide by **long division** (bring down one term at a time) or by **equating coefficients** (write $P(x) = D(x)Q(x) + R(x)$ with unknown coefficients in $Q$ and $R$, then match powers of $x$). Both give the same quotient and remainder.

**Worked example:** Divide $2x^3 - 3x^2 + 4x - 5$ by $(x - 2)$, stating the quotient and remainder.

1. Divide the leading terms: $2x^3 \div x = 2x^2$. Multiply back and subtract:
2. $$2x^3 - 3x^2 + 4x - 5 - 2x^2(x - 2) = x^2 + 4x - 5$$
3. Repeat: $x^2 \div x = x$. Subtract $x(x - 2) = x^2 - 2x$:
4. $$x^2 + 4x - 5 - (x^2 - 2x) = 6x - 5$$
5. Repeat: $6x \div x = 6$. Subtract $6(x - 2) = 6x - 12$:
6. $$6x - 5 - (6x - 12) = 7$$
7. The remainder $7$ has degree $0$, lower than $(x - 2)$, so stop:
8. $$2x^3 - 3x^2 + 4x - 5 = (x - 2)(2x^2 + x + 6) + 7$$

**Worked example:** Divide $x^3 + 2x^2 - x + 5$ by the quadratic $(x^2 + 1)$.

1. Leading terms: $x^3 \div x^2 = x$. Subtract $x(x^2 + 1) = x^3 + x$:
2. $$x^3 + 2x^2 - x + 5 - (x^3 + x) = 2x^2 - 2x + 5$$
3. Next: $2x^2 \div x^2 = 2$. Subtract $2(x^2 + 1) = 2x^2 + 2$:
4. $$2x^2 - 2x + 5 - (2x^2 + 2) = -2x + 3$$
5. The remainder $-2x + 3$ has degree $1 < 2$, so:
6. $$x^3 + 2x^2 - x + 5 = (x^2 + 1)(x + 2) + (-2x + 3)$$

> **Exam tip:** Keep a place for every power of $x$ when you set out the division — write $0x^2$ for a missing term so the columns line up.

## Factor and Remainder Theorems

**Remainder Theorem** — When a polynomial $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$. More generally, dividing by $(ax - b)$ leaves remainder $P\!\left(\frac{b}{a}\right)$.

*Notation:* P(x) \div (x - a) \ \text{leaves} \ P(a)

*Example:* $P(x) = x^2 + 3x + 1$ divided by $(x - 2)$ leaves $P(2) = 11$

**Factor Theorem** — $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$ (the special case of the remainder theorem when the remainder is $0$). For a factor $(ax - b)$, test $P\!\left(\frac{b}{a}\right) = 0$.

*Notation:* (x - a) \ \text{is a factor} \iff P(a) = 0

*Example:* $(x - 3)$ is a factor of $x^2 - x - 6$ because $P(3) = 0$

**Worked example:** When $P(x) = x^3 + 2x^2 + ax + 5$ is divided by $(x - 1)$ the remainder is $3$. Find $a$.

1. By the remainder theorem the remainder is $P(1)$, so set $P(1) = 3$:
2. $$1 + 2 + a + 5 = 3$$
3. Solve for $a$:
4. $$8 + a = 3 \implies a = -5$$

**Worked example:** Factorise $P(x) = x^3 - 3x^2 - x + 3$ completely.

1. Test factors of the constant term $3$. Try $x = 1$:
2. $$P(1) = 1 - 3 - 1 + 3 = 0$$
3. So $(x - 1)$ is a factor. Divide to find the quadratic factor:
4. $$x^3 - 3x^2 - x + 3 = (x - 1)(x^2 - 2x - 3)$$
5. Factorise the quadratic $x^2 - 2x - 3 = (x - 3)(x + 1)$:
6. $$P(x) = (x - 1)(x - 3)(x + 1)$$

> **Exam tip:** To factorise a cubic, test small values ($x = \pm 1, \pm 2, \pm 3, \ldots$ from the factors of the constant term) to spot one root, then divide out that factor and factorise the remaining quadratic.

## Proper and Improper Rational Functions

**Rational Function** — A ratio of two polynomials $P(x)$ (numerator) and non-zero $Q(x)$ (denominator). Classified by the degrees of the two polynomials.

*Notation:* f(x) = \frac{P(x)}{Q(x)}

*Example:* $\frac{3x^2 + 2x + 1}{x^2 - 4}$ is improper; $\frac{2x + 5}{x^2 - 4}$ is proper

Any improper rational function must first be simplified to a polynomial plus a proper rational function before partial fractions or expansion can be applied. This is done via polynomial long division or algebraic equating of coefficients.

**Worked example:** Simplify $\frac{3x^2 + 2x - 2}{x^2 + x - 2}$ into a polynomial plus a proper rational function.

1. Compare leading terms: $\frac{3x^2}{x^2} = 3$. Multiply the denominator by 3:
2. $$3(x^2 + x - 2) = 3x^2 + 3x - 6$$
3. Subtract this product from the original numerator:
4. $$(3x^2 + 2x - 2) - (3x^2 + 3x - 6) = -x + 4$$
5. Rewrite the original function:
6. $$\frac{3x^2 + 2x - 2}{x^2 + x - 2} = 3 + \frac{-x + 4}{x^2 + x - 2}$$

In the P3 partial-fractions syllabus the only improper case is $\deg(P) = \deg(Q)$ (equal degrees): dividing then gives a **constant** plus a proper fraction. A numerator of strictly higher degree, $\deg(P) > \deg(Q)$, does not arise in these questions.

## Partial Fraction Decomposition

Partial fraction decomposition breaks a proper rational function into a sum of simpler fractions, which is required for integration and series expansion. The form of the decomposition depends entirely on the factors of the denominator:

- Distinct linear factor $(ax + b)$: term $\frac{A}{ax + b}$
- Repeated linear factor $(ax + b)^2$: terms $\frac{A}{ax + b} + \frac{B}{(ax + b)^2}$
- Irreducible quadratic factor $(cx^2 + d)$: term $\frac{Ax + B}{cx^2 + d}$

**Worked example:** Decompose $\frac{-x + 4}{x^2 + x - 2}$ into partial fractions.

1. Factorise the denominator:
2. $$x^2 + x - 2 = (x + 2)(x - 1)$$
3. Set up the partial fraction form for two distinct linear factors:
4. $$\frac{-x + 4}{(x + 2)(x - 1)} = \frac{A}{x + 2} + \frac{B}{x - 1}$$
5. Multiply both sides by $(x + 2)(x - 1)$ to eliminate denominators:
6. $$-x + 4 = A(x - 1) + B(x + 2)$$
7. Substitute the roots of the denominator to find constants: $x = 1$ gives $3 = 3B \implies B = 1$; $x = -2$ gives $6 = -3A \implies A = -2$
8. Write the final decomposition:
9. $$\frac{-2}{x + 2} + \frac{1}{x - 1}$$

> **Exam tip:** Always verify your answer by combining partial fractions back into a single fraction to confirm it matches the original.

## Binomial Expansion for Negative and Fractional Powers

**General Binomial Expansion** — For rational $n$ and $|y| < 1$, the infinite binomial expansion is: $1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + ...$

*Notation:* (1 + y)^n

*Example:* For a positive integer $n$ the series terminates; for any other rational $n$ it is infinite and needs $|y| < 1$

To expand expressions of the form $(a + bx)^n$ where $a > 0$, first factor out $a^n$ to get the standard $(1 + y)^n$ form, then apply the general expansion. Always state the range of validity for the expansion.

**Worked example:** Find the first three non-zero terms of the expansion of $\frac{1}{1 + 2x}$ and state its range of validity.

1. Rewrite the function to match the standard form: $(1 + 2x)^{-1}$, so $n = -1$, $y = 2x$
2. Substitute into the general expansion up to the $x^2$ term:
3. $$(1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + ...$$
4. $$(1 + 2x)^{-1} = 1 + (-1)(2x) + \frac{(-1)(-2)}{2}(2x)^2 + ... = 1 - 2x + 4x^2 - ...$$
5. Calculate the range of validity from $|y| < 1$:
6. $$|2x| < 1 \implies |x| < \frac{1}{2}$$

## Common pitfalls

- **Wrong:** Skipping polynomial division for improper rational functions before partial fractions
  - Why it fails: Partial fraction rules only apply to proper fractions, leading to an incorrect decomposition
  - Correct: Always compare the degrees of numerator and denominator first; divide if $\deg(P) \geq \deg(Q)$
- **Wrong:** Only including the highest degree term for repeated linear factors
  - Why it fails: Missing lower degree terms changes the overall expression, leading to wrong constants
  - Correct: For $(ax + b)^k$, include terms from $\frac{A_1}{ax + b}$ up to $\frac{A_k}{(ax + b)^k}$
- **Wrong:** Forgetting to factor out the leading constant when expanding $(a + bx)^n$
  - Why it fails: The standard expansion is only valid for $(1 + y)^n$, so all coefficients will be wrong
  - Correct: Factor out $a^n$ first to get $a^n\left(1 + \frac{bx}{a}\right)^n$ before expanding
- **Wrong:** Omitting or incorrectly writing the range of validity for binomial expansion
  - Why it fails: Examiners consistently award at least one mark for the correct validity, which is required for all $n$ that is not a positive integer
  - Correct: Always rearrange $|y| < 1$ to get a condition on $x$ and write it clearly

## Cheatsheet

| Concept | Rule |
| --- | --- |
| Proper rational function | $\deg(P) < \deg(Q)$ |
| Improper rational function | $\deg(P) \geq \deg(Q)$; divide first |
| Distinct linear $(ax + b)$ | Term: $\frac{A}{ax + b}$ |
| Repeated linear $(ax + b)^2$ | Terms: $\frac{A}{ax + b} + \frac{B}{(ax + b)^2}$ |
| Irreducible quadratic $(cx^2+d)$ | Term: $\frac{Ax+B}{cx^2+d}$ |
| Binomial expansion $(1+y)^n$ | $1 + ny + \frac{n(n-1)}{2!}y^2 + ...$ |
| Expansion validity | $\|y\| < 1$ (rational $n$, not a positive integer) |

## What's next

This module lays the foundation for many core topics in CIE A-Level Pure Mathematics 3. Partial fraction decomposition is a required pre-processing step for integrating rational functions and solving differential equations with separable or linear forms. Binomial expansion of rational functions with negative and fractional powers is used for approximating functions, finding series expansions, and working with infinite sequences. Mastery of these algebraic manipulation skills is essential to access full marks on multi-step questions that combine multiple topics, which are common in P3 exams. Building on these skills will let you tackle more advanced concepts confidently.

- [Logarithms and Exponential Functions](https://www.owlsprep.com/study/cie-9709-u2-logarithms-and-exponential-functions/)
- [Trigonometry](https://www.owlsprep.com/study/cie-9709-u2-trigonometry/)
- [Differentiation](https://www.owlsprep.com/study/cie-9709-u2-differentiation/)

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