IGCSE Mathematics – Chapter 2: Algebra and Graphs

Cambridge IGCSE 0580 & Edexcel 4MA1 · Updated March 2026

Algebra is the language of mathematics. This chapter builds from basic notation and simplification all the way to quadratic equations, sequences, inequalities and the graphs of key function families. Master each skill in order — solving linear equations relies on expanding brackets, and quadratic methods rely on factorisation. The interactive Desmos figures will help you visualise how algebraic changes affect graphs.

Specification Note

Content labelled Extended is required for the Extended tier (Cambridge) or Higher tier (Edexcel) only. Core/Foundation tier students should still read these sections for enrichment.

1. Algebraic Notation and Simplification

1.1 Like Terms and Collecting

Algebra uses letters to represent unknown values or variables. Like terms have identical letter parts (including powers) and may be added or subtracted.

Example 1.1 — Collecting like terms

Simplify $5x^2 - 3x + 7 + 2x^2 + x - 4$.

Step 1 Group: $(5x^2 + 2x^2) + (-3x + x) + (7 - 4)$

Result $7x^2 - 2x + 3$

1.2 Substitution

Example 1.2 — Substitution

Given $p = 3$ and $q = -2$, evaluate $2p^2 - 3pq + q^2$.

$= 2(9) - 3(3)(-2) + (-2)^2 = 18 + 18 + 4 = 40$

1.3 Index Notation

$a^m \times a^n = a^{m+n}$; $\;(a^m)^n = a^{mn}$; $\;\dfrac{a^m}{a^n} = a^{m-n}$. Apply these rules when simplifying algebraic expressions.

Example 1.3 — Simplifying with indices

Simplify $\dfrac{6x^3y^2 \times 2xy^{-1}}{4x^2y^3}$.

Numerator: $12x^4y^1$. Denominator: $4x^2y^3$.
$= \dfrac{12}{4} \cdot x^{4-2} \cdot y^{1-3} = 3x^2y^{-2} = \dfrac{3x^2}{y^2}$

2. Expanding and Factorising

2.1 Single Brackets

Multiply every term inside the bracket by the term outside.

Example 2.1 — Expanding single brackets

Expand $3x(2x - 5y + 4)$.

$= 6x^2 - 15xy + 12x$

2.2 Double Brackets (FOIL)

Example 2.2 — Expanding double brackets

Expand $(3x - 2)(4x + 5)$.

First: $3x \times 4x = 12x^2$
Outer: $3x \times 5 = 15x$
Inner: $-2 \times 4x = -8x$
Last: $-2 \times 5 = -10$
$= 12x^2 + 7x - 10$

2.3 Special Products

Special Expansions

Difference of two squares: $(a + b)(a - b) = a^2 - b^2$
Perfect square (sum): $(a + b)^2 = a^2 + 2ab + b^2$
Perfect square (difference): $(a - b)^2 = a^2 - 2ab + b^2$

2.4 Factorising Quadratics

To factorise $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$.

Example 2.3 — Factorising quadratics

(a) Factorise $x^2 - 5x + 6$. (b) Factorise $2x^2 + 7x + 3$. (c) Factorise $9x^2 - 16$.

(a) Need two numbers with product $+6$ and sum $-5$: $(-2)$ and $(-3)$.
$= (x - 2)(x - 3)$

(b) $ac = 6$; need factors of $6$ adding to $7$: $1$ and $6$.
$2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) = (x + 3)(2x + 1)$

(c) Difference of two squares: $(3x)^2 - 4^2 = (3x-4)(3x+4)$

Practice 2A

(a) Expand and simplify $(2x + 3)^2 - (x - 1)(x + 4)$.
(b) Factorise completely $3x^2 - 75$.
(c) Factorise $6x^2 - 13x + 5$.

Show Solution

(a) $(4x^2 + 12x + 9) - (x^2 + 3x - 4) = 3x^2 + 9x + 13$

(b) $3(x^2 - 25) = 3(x-5)(x+5)$

(c) $ac = 30$; factors of $30$ summing to $-13$: $-3$ and $-10$.
$6x^2 - 3x - 10x + 5 = 3x(2x-1) - 5(2x-1) = (3x-5)(2x-1)$

3. Solving Linear Equations

Solving an equation means finding the value(s) of the unknown that make the equation true. The strategy is to isolate the unknown by performing the same operation on both sides.

Example 3.1 — Equations with fractions

Solve $\dfrac{2x - 1}{3} + \dfrac{x + 4}{2} = 5$.

Step 1 Multiply through by the LCM of the denominators (6):
$2(2x - 1) + 3(x + 4) = 30$

Step 2 Expand: $4x - 2 + 3x + 12 = 30$

Step 3 Collect: $7x + 10 = 30$

Step 4 $7x = 20 \Rightarrow x = \dfrac{20}{7}$

Example 3.2 — Equation with brackets on both sides

Solve $5(2x - 3) = 3(x + 4) - 1$.

$10x - 15 = 3x + 12 - 1$
$10x - 15 = 3x + 11$
$7x = 26$
$x = \dfrac{26}{7}$

4. Simultaneous Equations

4.1 Elimination Method

Example 4.1 — Elimination

Solve the system: $3x + 2y = 16$ and $5x - 2y = 8$.

Step 1 Add the equations (since $+2y$ and $-2y$ cancel):
$8x = 24 \Rightarrow x = 3$

Step 2 Substitute into first equation: $9 + 2y = 16 \Rightarrow 2y = 7 \Rightarrow y = 3.5$

Check $5(3) - 2(3.5) = 15 - 7 = 8$ ✓

4.2 Substitution Method

Example 4.2 — Substitution

Solve: $y = 3x - 1$ and $4x + 3y = 29$.

Substitute $y = 3x - 1$ into the second equation:
$4x + 3(3x - 1) = 29$
$4x + 9x - 3 = 29$
$13x = 32 \Rightarrow x = \dfrac{32}{13}$
$y = 3 \times \dfrac{32}{13} - 1 = \dfrac{96}{13} - \dfrac{13}{13} = \dfrac{83}{13}$

Practice 4A

Solve simultaneously: $2x + 5y = 1$ and $3x - 2y = 29$.

Show Solution

Multiply first equation by 2: $4x + 10y = 2$
Multiply second equation by 5: $15x - 10y = 145$
Add: $19x = 147 \Rightarrow x = \dfrac{147}{19}$... Let us instead use integers.
Multiply eq1 by 3: $6x + 15y = 3$; Multiply eq2 by 2: $6x - 4y = 58$
Subtract: $19y = -55 \Rightarrow y = -\dfrac{55}{19}$
Back-substitute: $2x = 1 - 5(-\dfrac{55}{19}) = 1 + \dfrac{275}{19} = \dfrac{294}{19} \Rightarrow x = \dfrac{147}{19}$

5. Quadratic Equations

5.1 Solving by Factorisation

Example 5.1 — Factorisation

Solve $x^2 - 3x - 10 = 0$.

$(x - 5)(x + 2) = 0 \Rightarrow x = 5$ or $x = -2$

5.2 The Quadratic Formula

Quadratic Formula

For $ax^2 + bx + c = 0$ (with $a \neq 0$): $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The expression $\Delta = b^2 - 4ac$ is called the discriminant. Extended

$\Delta > 0$: two distinct real roots
$\Delta = 0$: one repeated real root
$\Delta < 0$: no real roots

Example 5.2 — Quadratic formula

Solve $2x^2 - 5x - 1 = 0$, giving your answers to 2 decimal places.

$a = 2$, $b = -5$, $c = -1$
$\Delta = 25 + 8 = 33$
$x = \dfrac{5 \pm \sqrt{33}}{4}$
$x = \dfrac{5 + 5.745}{4} \approx 2.69$ or $x = \dfrac{5 - 5.745}{4} \approx -0.19$

5.3 Completing the Square Extended

Example 5.3 — Completing the square

Write $x^2 - 6x + 11$ in the form $(x - p)^2 + q$. Hence state the minimum value and the vertex.

$x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2$
Minimum value is $\mathbf{2}$, occurring at the vertex $(3, 2)$.

Practice 5A

(a) Solve $3x^2 + 2x - 8 = 0$ by factorisation.
(b) Solve $x^2 + 4x - 7 = 0$, giving your answers in surd form. Extended
(c) Find the value of $k$ such that $kx^2 - 4x + 1 = 0$ has exactly one solution. Extended

Show Solution

(a) $(3x - 4)(x + 2) = 0 \Rightarrow x = \dfrac{4}{3}$ or $x = -2$

(b) $x = \dfrac{-4 \pm \sqrt{16 + 28}}{2} = \dfrac{-4 \pm \sqrt{44}}{2} = \dfrac{-4 \pm 2\sqrt{11}}{2} = -2 \pm \sqrt{11}$

(c) Exactly one solution when $\Delta = 0$: $16 - 4k = 0 \Rightarrow k = 4$

Fig 1: Interactive quadratic $y = ax^2 + bx + c$. Use the sliders to change $a$, $b$ and $c$ and observe how the vertex, axis of symmetry and roots change.

6. Sequences and the nth Term

6.1 Term-to-Term and Position-to-Term Rules

Definitions

A term-to-term rule defines each term in relation to the previous term(s).
A position-to-term rule (nth term formula) gives the value of any term directly from its position number $n$.
An arithmetic sequence has a constant difference $d$ between consecutive terms. The nth term is $T_n = a + (n-1)d$, where $a$ is the first term.

Example 6.1 — Finding the nth term of an arithmetic sequence

Find the nth term of the sequence $7, 11, 15, 19, \ldots$

Step 1 Common difference $d = 4$.

Step 2 $T_n = a + (n-1)d = 7 + (n-1) \times 4 = 4n + 3$

Check $T_1 = 7$ ✓, $T_4 = 19$ ✓

Example 6.2 — Using the nth term

The nth term of a sequence is $T_n = 5n - 2$. Is 163 a term in this sequence?

Set $5n - 2 = 163 \Rightarrow 5n = 165 \Rightarrow n = 33$.
Since $n = 33$ is a positive integer, 163 is the 33rd term.

6.2 Quadratic Sequences Extended

A quadratic sequence has a constant second difference. If the second difference is $2k$, the nth term contains $kn^2$.

Example 6.3 — Quadratic sequence

Find the nth term of $3, 9, 19, 33, 51, \ldots$

1st differences: $6, 10, 14, 18$ (arithmetic sequence with $d = 4$)

2nd differences: $4, 4, 4$ — constant, so $kn^2$ where $2k = 4 \Rightarrow k = 2$.

Compare $T_n - 2n^2$: $3-2=1$, $9-8=1$, $19-18=1$. Constant $= 1$, so the sequence $T_n - 2n^2$ has nth term $1$.

Result $T_n = 2n^2 + 1$

Practice 6A

(a) The nth term of a sequence is $\dfrac{n^2 - 1}{n + 1}$. Show that this simplifies to $n - 1$ and state the type of sequence.
(b) Find the nth term of the quadratic sequence $4, 10, 20, 34, 52, \ldots$ Extended

Show Solution

(a) $\dfrac{n^2 - 1}{n + 1} = \dfrac{(n-1)(n+1)}{n+1} = n - 1$. This is a linear (arithmetic) sequence with first term $0$ and common difference $1$.

(b) 1st differences: $6, 10, 14, 18$. 2nd differences: $4, 4, 4$, so $k = 2$.
$T_n - 2n^2$: $4-2=2$, $10-8=2$. Constant $= 2$.
$T_n = 2n^2 + 2$

7. Algebraic Fractions and Inequalities

7.1 Solving Linear Inequalities

Key Rule

When multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality sign.

Example 7.1 — Solving and representing inequalities

Solve $3 - 2x \leq 11$ and represent the solution on a number line.

$-2x \leq 8$
Divide by $-2$ and reverse sign: $x \geq -4$

On a number line: solid circle at $-4$, arrow pointing right.

Example 7.2 — Integer solutions of a double inequality

Find all integers satisfying $-3 < 2x + 1 \leq 9$.

$-3 - 1 < 2x \leq 9 - 1 \Rightarrow -4 < 2x \leq 8 \Rightarrow -2 < x \leq 4$
Integer solutions: $\{-1, 0, 1, 2, 3, 4\}$

7.2 Algebraic Fractions Extended

Example 7.3 — Simplifying algebraic fractions

Simplify $\dfrac{x^2 - 9}{x^2 - x - 6}$.

Factorise: $\dfrac{(x-3)(x+3)}{(x-3)(x+2)} = \dfrac{x+3}{x+2}$ (valid for $x \neq 3$)

Example 7.4 — Adding algebraic fractions

Simplify $\dfrac{3}{x - 1} - \dfrac{2}{x + 2}$.

$= \dfrac{3(x+2) - 2(x-1)}{(x-1)(x+2)} = \dfrac{3x + 6 - 2x + 2}{(x-1)(x+2)} = \dfrac{x + 8}{(x-1)(x+2)}$

Practice 7A

(a) Solve $\dfrac{5x + 1}{2} > 3x - 4$ and list all positive integer solutions less than 10 that satisfy this inequality.
(b) Simplify $\dfrac{2x^2 + x - 3}{x^2 - 1}$. Extended
(c) Write $\dfrac{4}{x} + \dfrac{3}{2x - 1}$ as a single fraction. Extended

Show Solution

(a) $5x + 1 > 2(3x - 4) = 6x - 8 \Rightarrow 9 > x \Rightarrow x < 9$.
All real $x < 9$. Positive integer solutions less than 10: $\{1, 2, 3, 4, 5, 6, 7, 8\}$.

(b) $\dfrac{(2x + 3)(x - 1)}{(x-1)(x+1)} = \dfrac{2x + 3}{x + 1}$ (valid for $x \neq 1$)

(c) $\dfrac{4(2x-1) + 3x}{x(2x-1)} = \dfrac{8x - 4 + 3x}{x(2x-1)} = \dfrac{11x - 4}{x(2x-1)}$

8. Graphs of Functions

8.1 Linear Graphs

The equation $y = mx + c$ gives a straight line with gradient $m$ and $y$-intercept $c$. See Chapter 3 for full treatment.

8.2 Quadratic Graphs

$y = ax^2 + bx + c$ is a parabola. If $a > 0$ it opens upwards (U-shaped); if $a < 0$ it opens downwards (∩-shaped). The axis of symmetry is at $x = -\dfrac{b}{2a}$.

8.3 Cubic, Reciprocal and Exponential Graphs

Key Graph Shapes

Cubic $y = ax^3 + \ldots$: S-shaped curve with up to two turning points.
Reciprocal $y = \dfrac{k}{x}$: Two branches in opposite quadrants; asymptotes at $x = 0$ and $y = 0$.
Exponential $y = ab^x$ ($b > 0$, $b \neq 1$): If $b > 1$, growth curve. If $0 < b < 1$, decay curve. Always passes through $(0, a)$.

Fig 2: Comparison of key function types — linear (blue), quadratic (red), cubic (green) and reciprocal (purple) on the same axes.

Example 8.1 — Reading a graph

The graph of $y = x^2 - 4x + 3$ is drawn. Use the graph to write down: (a) the roots, (b) the vertex, (c) the axis of symmetry.

(a) Roots: where $y = 0$. Factorising: $(x-1)(x-3) = 0 \Rightarrow x = 1$ or $x = 3$. Roots at $(1, 0)$ and $(3, 0)$.

(b) Axis of symmetry: $x = \dfrac{1+3}{2} = 2$. Substitute: $y = 4 - 8 + 3 = -1$. Vertex at $(2, -1)$.

(c) Axis of symmetry: $x = 2$.

Practice 8A

(a) Sketch the graph of $y = 2x^2 - 8$, labelling the intercepts and vertex.
(b) State the equations of the asymptotes of $y = \dfrac{3}{x}$.
(c) The graph of $y = 2^x$ passes through $(0, 1)$. A second graph $y = 2^{x+3}$ is obtained. Describe the transformation. Extended

Show Solution

(a) $y$-intercept: $(0, -8)$. Roots: $x^2 = 4 \Rightarrow x = \pm 2$. Vertex: $(0, -8)$ (axis of symmetry $x = 0$).

(b) $x = 0$ and $y = 0$ (the coordinate axes).

(c) $2^{x+3} = 2^x \times 2^3 = 8 \times 2^x$. The graph is a translation 3 units to the left (or equivalently a vertical stretch by factor 8).

9. Mixed Practice Problems

Question 1

Simplify $\dfrac{x^2 - 4}{x^2 + 5x + 6}$ fully. Extended

Show Solution

$\dfrac{(x-2)(x+2)}{(x+2)(x+3)} = \dfrac{x-2}{x+3}$ (valid for $x \neq -2$)

Question 2

The sum of two numbers is $14$ and their product is $45$. Form and solve a quadratic equation to find the two numbers.

Show Solution

Let the numbers be $x$ and $14 - x$. Product: $x(14-x) = 45$
$14x - x^2 = 45 \Rightarrow x^2 - 14x + 45 = 0$
$(x - 5)(x - 9) = 0 \Rightarrow x = 5$ or $x = 9$.
The two numbers are 5 and 9.

Question 3

Find all integer values of $n$ such that $n^2 - 7n + 10 \leq 0$. Extended

Show Solution

Factorise: $(n-2)(n-5) \leq 0$. The parabola is $\leq 0$ between the roots.
$2 \leq n \leq 5$. Integer values: $n \in \{2, 3, 4, 5\}$.

Question 4

Solve simultaneously: $y = x^2 - 3x + 2$ and $y = x - 1$. Extended

Show Solution

$x^2 - 3x + 2 = x - 1 \Rightarrow x^2 - 4x + 3 = 0$
$(x-1)(x-3) = 0 \Rightarrow x = 1$ or $x = 3$
When $x=1$: $y = 0$. When $x = 3$: $y = 2$.
Solutions: $(1, 0)$ and $(3, 2)$.

Question 5

The nth term of a sequence is $T_n = 3n^2 - n$. Find the sum $T_1 + T_2 + T_3 + T_4$.

Show Solution

$T_1 = 2$, $T_2 = 10$, $T_3 = 24$, $T_4 = 44$.
Sum $= 2 + 10 + 24 + 44 = 80$.

Question 6

Solve $2x^2 - 3x - 4 = 0$, giving your answers to 3 significant figures.

Show Solution

$\Delta = 9 + 32 = 41$
$x = \dfrac{3 \pm \sqrt{41}}{4}$
$x \approx \dfrac{3 + 6.403}{4} \approx 2.35$ or $x \approx \dfrac{3 - 6.403}{4} \approx -0.851$

Question 7

Express $2x^2 + 12x + 5$ in the form $a(x + p)^2 + q$. Hence find the minimum value and the value of $x$ at which it occurs. Extended

Show Solution

$2(x^2 + 6x) + 5 = 2[(x + 3)^2 - 9] + 5 = 2(x + 3)^2 - 18 + 5 = 2(x + 3)^2 - 13$
Minimum value is $\mathbf{-13}$ at $x = -3$.

Question 8

A rectangle has length $(x + 7)$ cm and width $(x - 2)$ cm. Its area is $\text{40 cm}^2$. Find $x$.

Show Solution

$(x+7)(x-2) = 40$
$x^2 + 5x - 14 = 40$
$x^2 + 5x - 54 = 0$
$(x - ?)(x + ?)$: need factors of 54 differing by 5: $9$ and $-6$ give $+3$; $+9$ and $-6$? $9 \times 6 = 54$, $9 - 6 = 3$. Try discriminant: $\Delta = 25 + 216 = 241$.
$x = \dfrac{-5 + \sqrt{241}}{2} \approx \dfrac{-5 + 15.52}{2} \approx 5.26$ (take positive root since $x > 2$).

Question 9

The sequence $2, 5, 10, 17, 26, \ldots$ has nth term $T_n = n^2 + 1$. Which term of the sequence is equal to $290$?

Show Solution

$n^2 + 1 = 290 \Rightarrow n^2 = 289 \Rightarrow n = 17$.
$290$ is the 17th term.

Question 10

Solve $\dfrac{3}{x + 1} + \dfrac{2}{x - 2} = 1$. Extended

Show Solution

Multiply through by $(x+1)(x-2)$:
$3(x-2) + 2(x+1) = (x+1)(x-2)$
$3x - 6 + 2x + 2 = x^2 - x - 2$
$5x - 4 = x^2 - x - 2$
$x^2 - 6x + 2 = 0$
$x = \dfrac{6 \pm \sqrt{36 - 8}}{2} = \dfrac{6 \pm \sqrt{28}}{2} = 3 \pm \sqrt{7}$