Algebra

CIE A-Level Mathematics· Pure 2: 2.1 Algebra; Pure 3: 3.1 Algebra· 5 min read

1. The Modulus Function★★☆☆☆P2 & P3 only⏱ 15 min

📘 Definition

Modulus (Absolute Value)

|x|

The modulus $∣ x ∣$ is the magnitude of $x$ : $∣ x ∣ = x$ if $x \geq 0$ and $∣ x ∣ = - x$ if $x < 0$ . It is never negative and measures the distance of $x$ from $0$ on the number line.

Example:

$∣5∣ = 5$ , $∣ - 5∣ = 5$ , and $∣ x - 3∣$ is the distance between $x$ and $3$

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

📘 Definition

Key Modulus Relations

|a| = |b| \iff a^2 = b^2

Two useful equivalences: $∣ a ∣ = ∣ b ∣ ⟺ a^{2} = b^{2}$ (true for all $a$ and $b$ ), which lets you remove modulus signs by squaring; and, for $b > 0$ , $∣ x - a ∣ < b ⟺ a - b < x < a + b$ , which turns a modulus inequality into a double inequality.

Example:

$∣ x - 4∣ < 3 ⟺ 1 < x < 7$

📐 Worked Example

Solve $∣3 x - 2∣ = ∣2 x + 7∣$ .

1
Both sides are moduli, so square both sides to remove them (using $∣ a ∣ = ∣ b ∣ ⟺ a^{2} = b^{2}$ ):
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 $∣2 x - 1∣ < 5$ .

1
A single modulus less than a positive number becomes a double inequality, $- 5 < 2 x - 1 < 5$ :
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 $\pm$ .

2. Polynomial Division★★★☆☆P2 & P3 only⏱ 15 min

📘 Definition

Division Identity

P(x) = D(x)\,Q(x) + R(x)

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 $de g R < de g D$ . 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 $p x + q$ .

Example:

$x^{2} - 5 x + 6 = (x - 2) (x - 3) + 0$ , 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 $2 x^{3} - 3 x^{2} + 4 x - 5$ by $(x - 2)$ , stating the quotient and remainder.

1
Divide the leading terms: $2 x^{3} \div x = 2 x^{2}$ . 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 $x (x - 2) = x^{2} - 2 x$ :
4
$x^2 + 4x - 5 - (x^2 - 2x) = 6x - 5$
5
Repeat: $6 x \div x = 6$ . Subtract $6 (x - 2) = 6 x - 12$ :
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 $x^{3} + 2 x^{2} - x + 5$ 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: $2 x^{2} \div x^{2} = 2$ . Subtract $2 (x^{2} + 1) = 2 x^{2} + 2$ :
4
$2x^2 - 2x + 5 - (2x^2 + 2) = -2x + 3$
5
The remainder $- 2 x + 3$ 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 $0 x^{2}$ for a missing term so the columns line up.

3. Factor and Remainder Theorems★★★☆☆P2 & P3 only⏱ 15 min

📘 Definition

Remainder Theorem

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

When a polynomial $P (x)$ is divided by $(x - a)$ , the remainder is $P (a)$ . More generally, dividing by $(a x - b)$ leaves remainder $P (\frac{b}{a})$ .

Example:

$P (x) = x^{2} + 3 x + 1$ divided by $(x - 2)$ leaves $P (2) = 11$

📘 Definition

Factor Theorem

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

$(x - a)$ 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 $(a x - b)$ , test $P (\frac{b}{a}) = 0$ .

Example:

$(x - 3)$ is a factor of $x^{2} - x - 6$ because $P (3) = 0$

📐 Worked Example

When $P (x) = x^{3} + 2 x^{2} + a x + 5$ 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 $P (x) = x^{3} - 3 x^{2} - x + 3$ 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 $x^{2} - 2 x - 3 = (x - 3) (x + 1)$ :
6
$P(x) = (x - 1)(x - 3)(x + 1)$

Exam tip:

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

4. Proper and Improper Rational Functions★★☆☆☆P3 only⏱ 10 min

📘 Definition

Rational Function

f(x) = \frac{P(x)}{Q(x)}

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

Example:

$\frac{3 x ^{2} + 2 x + 1}{x ^{2} - 4}$ is improper; $\frac{2 x + 5}{x ^{2} - 4}$ 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 $\frac{3 x ^{2} + 2 x - 2}{x ^{2} + x - 2}$ into a polynomial plus a proper rational function.

1
Compare leading terms: $\frac{3 x ^{2}}{x ^{2}} = 3$ . 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 $de g (P) = de g (Q)$ (equal degrees): dividing then gives a constant plus a proper fraction. A numerator of strictly higher degree, $de g (P) > de g (Q)$ , does not arise in these questions.

5. Partial Fraction Decomposition★★★☆☆P3 only⏱ 20 min

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 $(a x + b)$ : term $\frac{A}{a x + b}$
Repeated linear factor $(a x + b)^{2}$ : terms $\frac{A}{a x + b} + \frac{B}{( a x + b ) ^{2}}$
Irreducible quadratic factor $(c x^{2} + d)$ : term $\frac{A x + B}{c x ^{2} + d}$

📐 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 $3 = 3 B ⟹ B = 1$ ; $x = - 2$ gives $6 = - 3 A ⟹ A = - 2$
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.

6. Binomial Expansion for Negative and Fractional Powers★★★★☆P3 only⏱ 20 min

📘 Definition

General Binomial Expansion

(1 + y)^n

For rational $n$ and $∣ y ∣ < 1$ , the infinite binomial expansion is: $1 + n y + \frac{n ( n - 1 )}{2 !} y^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} y^{3} + ...$

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 $(a + b x)^{n}$ 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 $\frac{1}{1 + 2 x}$ and state its range of validity.

1
Rewrite the function to match the standard form: $(1 + 2 x)^{- 1}$ , so $n = - 1$ , $y = 2 x$
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
$|2x| < 1 \implies |x| < \frac{1}{2}$

7. Common Pitfalls

Wrong move:

Skipping polynomial division for improper rational functions before partial fractions

Why:

Partial fraction rules only apply to proper fractions, leading to an incorrect decomposition

Correct move:

Always compare the degrees of numerator and denominator first; divide if $de g (P) \geq de g (Q)$

Wrong move:

Only including the highest degree term for repeated linear factors

Why:

Missing lower degree terms changes the overall expression, leading to wrong constants

Correct move:

For $(a x + b)^{k}$ , include terms from $\frac{A _{1}}{a x + b}$ up to $\frac{A _{k}}{( a x + b ) ^{k}}$

Wrong move:

Forgetting to factor out the leading constant when expanding $(a + b x)^{n}$

Why:

The standard expansion is only valid for $(1 + y)^{n}$ , so all coefficients will be wrong

Correct move:

Factor out $a^{n}$ first to get $a^{n} (1 + \frac{b x}{a})^{n}$ before expanding

Wrong move:

Omitting or incorrectly writing the range of validity for binomial expansion

Why:

Examiners consistently award at least one mark for the correct validity, which is required for all $n$ that is not a positive integer

Correct move:

Always rearrange $∣ y ∣ < 1$ to get a condition on $x$ and write it clearly

8. Quick Reference Cheatsheet

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

Going deeper

syllabusCIE 9709 A-Level Mathematics Syllabus

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.