GRADE 12 / MATHEMATICS
STUDY SMARTER.
SOLVE BETTER.
Supplementary Study Notes
IEB EDITION
Clear topic explanations
Step-by-step worked examples
Practice questions & explained memos
12 focused chapters
Independent learning resource • Not an official DBE or IEB publication
,YOUR STUDY ROADMAP
CONTENTS
Jump to a topic, work through the examples, then test yourself.
01 Algebra, Equations & Inequalities 3
02 Sequences & Series 13
03 Functions, Graphs & Inverses 17
04 Exponents & Logarithms 21
05 Financial Mathematics 25
06 Differential Calculus 29
07 Applications of Calculus 33
08 Counting & Probability 37
09 Analytical Geometry 41
10 Trigonometry 45
11 Euclidean Geometry 49
12 Statistics & Regression 53
Page numbers refer to this newly formatted PDF.
Study tip: Try the questions before checking the memorandums.
, How to use this chapter
Read a lesson before looking at its examples. Cover the next line of each solution and predict what
happens. Attempt the practice questions without the memorandum. Afterwards, check both the answer
and the reasoning; redo any question you could not explain.
Learning outcomes
• Manipulate algebraic expressions and factorise correctly.
• Solve linear, quadratic and simultaneous equations.
• Solve quadratic inequalities with a sign chart.
• Work with surds, fractional exponents and restrictions.
• Use the discriminant to reason about the nature of roots.
• Model a word problem and interpret the solution in context.
Lesson 1 — Algebraic language and factorisation
An expression has no equals sign, while an equation asserts that two expressions are equal. Factorisation
rewrites a sum or difference as a product. It is the reverse of expanding brackets. Factorisation is
particularly useful when a product is equal to zero.
Core patterns
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
a² − b² = (a − b)(a + b)
For a quadratic ax² + bx + c, look for two numbers whose product is ac and whose sum is b. Split the
middle term and group.
Worked example 1.1 — Factorise completely
6x² − x − 2
Step 1: Multiply the first and last coefficients: 6 × (−2) = −12.
Step 2: Find two integers with product −12 and sum −1: 3 and −4.
6x² + 3x − 4x − 2
Step 3: Group and factor each pair.
3x(2x + 1) − 2(2x + 1)
= (3x − 2)(2x + 1)
Check: expanding gives 6x² + 3x − 4x − 2 = 6x² − x − 2.
Worked example 1.2 — Difference of squares
9p² − 49 = (3p − 7)(3p + 7)
Independent Grade 12 Mathematics Learning Series | Chapter 1 2
, Common error: a² + b² is not a difference of squares. You cannot factor it into (a − b)(a + b) over the real
numbers.
Lesson 2 — Equations and the zero-product rule
The zero-product rule says that if A × B = 0, then A = 0 or B = 0. It applies only when the product equals
zero. Always rearrange a quadratic equation into the form ax² + bx + c = 0 before factorising.
Worked example 2.1 — Factorisation
2x² − 7x + 3 = 0
(2x − 1)(x − 3) = 0
2x − 1 = 0 or x − 3 = 0
x = 1/2 or x = 3
Check both values by substitution into the original equation.
Worked example 2.2 — Quadratic formula
For ax² + bx + c = 0, with a ≠ 0:
x = [−b ± √(b² − 4ac)] / (2a)
3x² + 2x − 4 = 0
Identify a = 3, b = 2, c = −4. The discriminant is 2² − 4(3)(−4) = 52.
x = [−2 ± √52] / 6 = [−1 ± √13] / 3
Use exact surd form unless the question requests decimals.
Worked example 2.3 — Equations with fractions
2/(x − 1) + 1 = 5/(x − 1)
First state the restriction: x ≠ 1. Multiply every term by x − 1.
2 + (x − 1) = 5 ⇒ x = 4
Since 4 ≠ 1, the solution is valid. Multiplying by a denominator without stating restrictions can introduce
invalid answers.
Lesson 3 — Simultaneous equations
A simultaneous solution is a pair (x, y) satisfying both equations. Use substitution when one variable is
already isolated. Use elimination when coefficients are easy to match.
Worked example 3.1 — A line and a parabola
y = x + 2 and y = x² − 4
Because both expressions equal y, set them equal.
x + 2 = x² − 4 ⇒ x² − x − 6 = 0
Independent Grade 12 Mathematics Learning Series | Chapter 1 3
STUDY SMARTER.
SOLVE BETTER.
Supplementary Study Notes
IEB EDITION
Clear topic explanations
Step-by-step worked examples
Practice questions & explained memos
12 focused chapters
Independent learning resource • Not an official DBE or IEB publication
,YOUR STUDY ROADMAP
CONTENTS
Jump to a topic, work through the examples, then test yourself.
01 Algebra, Equations & Inequalities 3
02 Sequences & Series 13
03 Functions, Graphs & Inverses 17
04 Exponents & Logarithms 21
05 Financial Mathematics 25
06 Differential Calculus 29
07 Applications of Calculus 33
08 Counting & Probability 37
09 Analytical Geometry 41
10 Trigonometry 45
11 Euclidean Geometry 49
12 Statistics & Regression 53
Page numbers refer to this newly formatted PDF.
Study tip: Try the questions before checking the memorandums.
, How to use this chapter
Read a lesson before looking at its examples. Cover the next line of each solution and predict what
happens. Attempt the practice questions without the memorandum. Afterwards, check both the answer
and the reasoning; redo any question you could not explain.
Learning outcomes
• Manipulate algebraic expressions and factorise correctly.
• Solve linear, quadratic and simultaneous equations.
• Solve quadratic inequalities with a sign chart.
• Work with surds, fractional exponents and restrictions.
• Use the discriminant to reason about the nature of roots.
• Model a word problem and interpret the solution in context.
Lesson 1 — Algebraic language and factorisation
An expression has no equals sign, while an equation asserts that two expressions are equal. Factorisation
rewrites a sum or difference as a product. It is the reverse of expanding brackets. Factorisation is
particularly useful when a product is equal to zero.
Core patterns
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
a² − b² = (a − b)(a + b)
For a quadratic ax² + bx + c, look for two numbers whose product is ac and whose sum is b. Split the
middle term and group.
Worked example 1.1 — Factorise completely
6x² − x − 2
Step 1: Multiply the first and last coefficients: 6 × (−2) = −12.
Step 2: Find two integers with product −12 and sum −1: 3 and −4.
6x² + 3x − 4x − 2
Step 3: Group and factor each pair.
3x(2x + 1) − 2(2x + 1)
= (3x − 2)(2x + 1)
Check: expanding gives 6x² + 3x − 4x − 2 = 6x² − x − 2.
Worked example 1.2 — Difference of squares
9p² − 49 = (3p − 7)(3p + 7)
Independent Grade 12 Mathematics Learning Series | Chapter 1 2
, Common error: a² + b² is not a difference of squares. You cannot factor it into (a − b)(a + b) over the real
numbers.
Lesson 2 — Equations and the zero-product rule
The zero-product rule says that if A × B = 0, then A = 0 or B = 0. It applies only when the product equals
zero. Always rearrange a quadratic equation into the form ax² + bx + c = 0 before factorising.
Worked example 2.1 — Factorisation
2x² − 7x + 3 = 0
(2x − 1)(x − 3) = 0
2x − 1 = 0 or x − 3 = 0
x = 1/2 or x = 3
Check both values by substitution into the original equation.
Worked example 2.2 — Quadratic formula
For ax² + bx + c = 0, with a ≠ 0:
x = [−b ± √(b² − 4ac)] / (2a)
3x² + 2x − 4 = 0
Identify a = 3, b = 2, c = −4. The discriminant is 2² − 4(3)(−4) = 52.
x = [−2 ± √52] / 6 = [−1 ± √13] / 3
Use exact surd form unless the question requests decimals.
Worked example 2.3 — Equations with fractions
2/(x − 1) + 1 = 5/(x − 1)
First state the restriction: x ≠ 1. Multiply every term by x − 1.
2 + (x − 1) = 5 ⇒ x = 4
Since 4 ≠ 1, the solution is valid. Multiplying by a denominator without stating restrictions can introduce
invalid answers.
Lesson 3 — Simultaneous equations
A simultaneous solution is a pair (x, y) satisfying both equations. Use substitution when one variable is
already isolated. Use elimination when coefficients are easy to match.
Worked example 3.1 — A line and a parabola
y = x + 2 and y = x² − 4
Because both expressions equal y, set them equal.
x + 2 = x² − 4 ⇒ x² − x − 6 = 0
Independent Grade 12 Mathematics Learning Series | Chapter 1 3
