Learning objectives
- I can explain and apply: Distributivity and identities.
- I can explain and apply: Factorisation.
- I can explain and apply: Restrictions on rational expressions.
Greek Gymnasium Grade 3 / Mathematics / Two-unit revision pilot
Algebraic expressions
Algebraic expressions: structured theory, worked examples, answered practice, and a mastery checklist for Greek Gymnasium Grade 3.
CHAPTER PLAN
Learn, practise, and check your progress
Follow the steps in order or jump directly to the part you need.
—/5
completed
Estimated active study time
156 minutes
1. Understand
Objectives, key ideas, and structured theory.
Open →2. Follow the method
Worked examples that explain every step.
Open →3. Practise
Graded tasks with hints and answer guidance.
Open →4. Consolidate
Assignments from core work to challenge.
Open →5. Check
A timed, marked test with solution guidance.
Open →
Progress is stored only in this browser and does not require AI credits.
Key ideas and checks
- (x±a)²=x²±2ax+a².
- x²−a²=(x−a)(x+a).
- Cancel factors, not terms in sums.
Unit
Core theory
The essential chapter ideas in a clear sequence before practice.
Understanding the concept
Under distributivity, multiply every term of the first bracket by every term of the second. Collect only like terms: x and x² are not terms of the same kind.
Method and conditions
Factorisation writes a sum as a product. Check for a common factor before using an identity. Expanding the product must reproduce the original expression exactly.
Checking and coverage boundaries
For a rational expression, record where the original denominator is zero before simplifying. Cancelling a common factor does not restore excluded values. For example, (x²−9)/(x−3)=x+3 holds only for x≠3.
Original pilot material — human educator review pending
Written specifically for YourFavTeacher with AI assistance, without copying past-paper prompts. Technical and numerical checks do not replace review by a human educator. It is not an official paper or endorsed by an examination body.
Mathematics
Worked examples
Follow the method step by step and check why every step is valid.
Worked example 1
Expand and simplify (6x+2)(x−5).
- 1Multiply every term in the first bracket by every term in the second.
- 26x²−30x+2x−10
- 3At x=0 both forms give −10.
6x²-28x−10
Worked example 2
Simplify (x²−25)/(x−5) and state the restriction.
- 1The original denominator requires x≠5.
- 2x²−25=(x−5)(x+5)
- 3Cancel the common factor only for x≠5.
x+5, x≠5
Worked example 3
Simplify 6(x+2)−5(x−1).
- 1Watch the negative sign before the second bracket.
- 26x+12−5x+5
- 3The coefficient of x is 6−5=1.
x+17
Algebraic expressions
Practice with answers
Eight graded tasks from core fluency to exam-style application. Work independently before opening a hint or answer.
Core fluency6 minutes
Expand and simplify (3x+2)(x−2).
Hint
Multiply every term in the first bracket by every term in the second.
Answer guide
Multiply every term in the first bracket by every term in the second. 3x²−6x+2x−4 At x=0 both forms give −4. 3x²-4x−4
Core fluency6 minutes
Expand (x+3)² and explain the middle term.
Hint
A square is a product of two identical brackets.
Answer guide
A square is a product of two identical brackets. (x+3)(x+3)=x²+3x+3x+9 The two cross terms add; they must not be omitted. x²+6x+9
Core fluency6 minutes
Factor x²−9.
Hint
Recognise a difference of two squares.
Answer guide
Recognise a difference of two squares. x²−3²=(x−3)(x+3) Expanding makes the cross terms cancel. (x−3)(x+3)
Application6 minutes
Factor 3x²+6x.
Hint
Both terms have a common factor.
Answer guide
Both terms have a common factor. 3x²+6x=3x(x+2) Expand to check both original terms. 3x(x+2)
Application6 minutes
Simplify (x²−4)/(x−2) and state the restriction.
Hint
The original denominator requires x≠2.
Answer guide
The original denominator requires x≠2. x²−4=(x−2)(x+2) Cancel the common factor only for x≠2. x+2, x≠2
Application6 minutes
Calculate (3x²)·(−2x³).
Hint
Multiply coefficients and add exponents of the same base.
Answer guide
Multiply coefficients and add exponents of the same base. 3·(−2)=−6; x²·x³=x⁵ The sign is negative, not positive. −6x⁵
Reasoning6 minutes
Someone writes (x−2)²=x²−4. Identify the error by expanding.
Hint
(x−2)(x−2)=x²−2x−2x+4
Answer guide
(x−2)(x−2)=x²−2x−2x+4 There is a middle term and the constant term is positive. At x=0 the square is 4, not −4. x²−4x+4
Reasoning6 minutes
Simplify 3(x+2)−2(x−1).
Hint
Watch the negative sign before the second bracket.
Answer guide
Watch the negative sign before the second bracket. 3x+6−2x+2 The coefficient of x is 3−2=1. x+8
Common mistakes
- Omitting the middle term in a squared binomial.
- Cancelling individual terms inside a sum.
- Losing the original denominator restriction.
Mastery check
- I can solve independently and check: Distributivity and identities.
- I can solve independently and check: Factorisation.
- I can solve independently and check: Restrictions on rational expressions.
Algebraic expressions
Chapter homework
Six distinct assignments from core fluency to challenge, each with an estimated time, hint, and answer guide.
Core8 minutes
Expand and simplify (4x+2)(x−3).
Hint
Multiply every term in the first bracket by every term in the second.
Answer guide
Multiply every term in the first bracket by every term in the second. 4x²−12x+2x−6 At x=0 both forms give −6. 4x²-10x−6
Core8 minutes
Expand (x+4)² and explain the middle term.
Hint
A square is a product of two identical brackets.
Answer guide
A square is a product of two identical brackets. (x+4)(x+4)=x²+4x+4x+16 The two cross terms add; they must not be omitted. x²+8x+16
Stretch8 minutes
Factor x²−16.
Hint
Recognise a difference of two squares.
Answer guide
Recognise a difference of two squares. x²−4²=(x−4)(x+4) Expanding makes the cross terms cancel. (x−4)(x+4)
Stretch8 minutes
Factor 4x²+12x.
Hint
Both terms have a common factor.
Answer guide
Both terms have a common factor. 4x²+12x=4x(x+3) Expand to check both original terms. 4x(x+3)
Challenge8 minutes
Simplify (x²−9)/(x−3) and state the restriction.
Hint
The original denominator requires x≠3.
Answer guide
The original denominator requires x≠3. x²−9=(x−3)(x+3) Cancel the common factor only for x≠3. x+3, x≠3
Challenge8 minutes
Calculate (4x²)·(−3x³).
Hint
Multiply coefficients and add exponents of the same base.
Answer guide
Multiply coefficients and add exponents of the same base. 4·(−3)=−12; x²·x³=x⁵ The sign is negative, not positive. −12x⁵
45 minutes / 40 marks
Full chapter test
A timed, full-mark self-assessment with model-answer guidance.
Show working, units and a final check. Equivalent correct methods are accepted. Timing is a revision guide, not official examination conditions.
1. Expand and simplify (5x+2)(x−4).
5 marksAnswer guide
Multiply every term in the first bracket by every term in the second. 5x²−20x+2x−8 At x=0 both forms give −8. 5x²-18x−8
2. Expand (x+5)² and explain the middle term.
5 marksAnswer guide
A square is a product of two identical brackets. (x+5)(x+5)=x²+5x+5x+25 The two cross terms add; they must not be omitted. x²+10x+25
3. Factor x²−25.
5 marksAnswer guide
Recognise a difference of two squares. x²−5²=(x−5)(x+5) Expanding makes the cross terms cancel. (x−5)(x+5)
4. Factor 5x²+20x.
5 marksAnswer guide
Both terms have a common factor. 5x²+20x=5x(x+4) Expand to check both original terms. 5x(x+4)
5. Simplify (x²−16)/(x−4) and state the restriction.
5 marksAnswer guide
The original denominator requires x≠4. x²−16=(x−4)(x+4) Cancel the common factor only for x≠4. x+4, x≠4
6. Calculate (5x²)·(−4x³).
5 marksAnswer guide
Multiply coefficients and add exponents of the same base. 5·(−4)=−20; x²·x³=x⁵ The sign is negative, not positive. −20x⁵
7. Someone writes (x−4)²=x²−16. Identify the error by expanding.
5 marksAnswer guide
(x−4)(x−4)=x²−4x−4x+16 There is a middle term and the constant term is positive. At x=0 the square is 16, not −16. x²−8x+16
8. Simplify 5(x+2)−4(x−1).
5 marksAnswer guide
Watch the negative sign before the second bracket. 5x+10−4x+4 The coefficient of x is 5−4=1. x+14
Unit
Official sources and verification
Curriculum reference sources. Always confirm the teaching sequence with the school and tutor.
Mathematics
A bridge year into Lyceum: students need confidence in algebraic manipulation and basic proof thinking.
What this chapter covers
The structure follows the official textbook layout and is used to organise study.
Distributivity and identities
Factorisation
Restrictions on rational expressions
Where to focus
The areas that usually create mistakes or need extra revision.
I can solve independently and check: Distributivity and identities.
I can solve independently and check: Factorisation.
I can solve independently and check: Restrictions on rational expressions.
Sources, daily material, and resources
Where to start: textbook, daily material, PDFs, videos, and worked examples.
Start from the official textbook or specification referenced on the subject page.
Use the notes and examples as support, not as a replacement for the official syllabus.
Check the current syllabus version before exam preparation.
Practice by subtopic
Targeted practice before full tests so coverage is clear.
Expand and simplify (3x+2)(x−2).
Expand (x+3)² and explain the middle term.
Factor x²−9.
Factor 3x²+6x.
Simplify (x²−4)/(x−2) and state the restriction.
Calculate (3x²)·(−2x³).
Someone writes (x−2)²=x²−4. Identify the error by expanding.
Simplify 3(x+2)−2(x−1).
Mocks and progress checks
How to measure progress in this chapter and when it enters a cumulative mock.
Start with untimed practice by subtopic.
Move to a short timed checkpoint only after completing the mastery checklist.
Record each error with the correct method and revisit it after 48 hours.
Next step
What to do after finishing the chapter and how it connects to the next unit.
Complete the practice without support.
Explain the core method aloud in under two minutes.
Continue to the next chapter or request targeted tutor support.
Note: for the official examinable syllabus of each school year, always confirm with the school, tutor, and current Ministry/IEP announcements.