Secondary 2 Mathematics Classroom | Chapter 2: Algebraic Expansion, Formulae and Identities | G2/G3

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SECONDARY 2 MATHEMATICS CLASSROOM · CHAPTER 2 · ALGEBRAIC EXPANSION · FORMULAE · IDENTITIES · G2/G3

Algebraic Expansion, Formulae and Identities: Change the Form Without Changing the Mathematics

The central skill in this classroom is not “opening brackets”. It is preserving equivalence while the written form changes.

Secondary 2 algebra asks you to do something more demanding than calculate with letters. You must read structure, recognise terms and factors, distribute multiplication correctly, collect like terms, substitute complete values into expressions and formulae, rearrange relationships, recognise identities and decide whether two different-looking forms are actually equivalent.

Classroom rule: read the structure → identify the operation → transform one justified step at a time → preserve every sign and bracket → verify by expansion, substitution or reverse operation.

The exact depth of algebraic manipulation varies by subject level and school sequence. This classroom teaches the common Secondary 2 operating system and marks more demanding multi-step work as a bridge where appropriate. Chapter 3 will own factorisation and algebraic fractions in greater depth; this chapter builds the expansion, equivalence, formula and identity control those topics depend on.

Official reference: MOE G2 and G3 Mathematics Syllabuses.

Navigate: retrieval · structure · single brackets · double brackets · formulae · changing the subject · identities · verification · misconception clinic · guided practice · assessment transfer · exit ticket.

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Featured Answer: What Is Algebraic Expansion?

Expansion uses the distributive law to rewrite a product as an equivalent sum. For example, 3(x+4)=3x+12. The form changes; the value does not. Factorisation later reverses that process.

Expansion does not create a new expression. It reveals the same expression in another form.

The Simple Classroom Answer

• Expression: a mathematical object such as 3x+5.
• Term: a part separated by addition or subtraction.
• Factor: a part joined by multiplication.
• Expansion: distribute multiplication across a sum or difference.
• Equivalent expressions: different forms with the same value for every permitted input.
• Formula: an equation expressing a relationship among quantities.
• Subject: the variable isolated on one side of a formula.
• Identity: an equation true for every permitted value of its variable.

How to Use This Classroom

1. Circle brackets before expanding.
2. Read the sign attached to every term.
3. Write each product explicitly while learning.
4. Collect only like terms.
5. Use brackets when substituting negative numbers.
6. When changing the subject, apply the same inverse operation to both sides.
7. Protect complete numerators and denominators.
8. Test a proposed identity by valid algebra; use numerical substitution only as a quick error detector.
9. Re-expand a factorised form or substitute a value to verify equivalence.
10. Return later with the method label removed.

1. Retrieval: Algebra Is Compressed Arithmetic

3x means 3×x. x² means x×x. 2(x+5) means two copies of the entire quantity x+5.

2. An Expression Is Not an Equation

3x+5 is an expression. 3x+5=20 is an equation. The expression can be simplified or evaluated; the equation can be solved.

3. Like Terms Represent the Same Algebraic Unit

3x+5x=8x because both terms are multiples of x. But 3x+5x² cannot be collected because x and x² are different algebraic quantities.

4. Negative Signs Belong to Terms

In 4x²−7x+3, the terms are 4x², −7x and 3. The negative sign is part of the middle term.

5. A Quick Retrieval Diagnostic

1. Simplify 4x+7x.
2. Decide whether 5x+2x² can be collected.
3. Expand 3(x+2).
4. Expand −2(x−5).
5. Evaluate x²+3x when x=−2.

Answers

11x. No. 3x+6. −2x+10. 4−6=−2.

6. Read Terms and Factors Before Choosing an Operation

In 6x+12, the terms are 6x and 12. In 6(x+2), the factors are 6 and x+2. The two expressions are equivalent, but their visible structures are different.

7. Addition and Multiplication Permit Different Moves

Like terms may be added. Common factors may be extracted. A letter cannot be “cancelled” across addition simply because it appears in two places.

8. Equivalent Expressions Give the Same Value for Every Permitted Input

2(x+3) and 2x+6 are equivalent. At x=0, both are 6. At x=5, both are 16. The algebraic reason is distribution, not the fact that two test values happened to agree.

9. Teacher Model 1: Read the Job Before Acting

For 6x+12:

• “simplify” → already simplified as a sum;
• “factorise” → 6(x+2);
• “evaluate at x=3” → 30;
• “solve 6x+12=30” → x=3.

The same symbols can sit inside different mathematical jobs.

10. The Equals Sign Is a Claim of Equal Value

Do not write 3+5=8×2=16. The chain claims 8=16. Each equals sign must connect expressions with the same value.

11. The Distributive Law Controls Single-Bracket Expansion

a(b+c)=ab+ac.

12. Every Term Inside the Bracket Must Be Multiplied

4(x+3)=4x+12, not 4x+3.

13. Teacher Model 2: Positive Multiplier

Expand 5(2x+7).

5×2x=10x and 5×7=35.

Answer: 10x+35.

14. Distribution Over Subtraction Preserves the Signed Term

3(x−4)=3x−12 because the second term is −4.

15. Teacher Model 3: Negative Multiplier

Expand −3(2x−5).

−3×2x=−6x and −3×−5=+15.

Answer: −6x+15.

16. A Minus Sign Before a Bracket Is Multiplication by −1

−(x−7)=−x+7.

This makes sign reversal structural rather than mnemonic.

17. Expand Before Collecting Like Terms

2(x+3)+5x=2x+6+5x=7x+6.

18. Teacher Model 4: Two Single Brackets

Simplify 3(x+4)−2(x−5).

3x+12−2x+10=x+22.

The second bracket produces +10 because −2×−5=+10.

19. Teacher Model 5: Coefficients and Several Terms

Simplify 4(2a−3b)+5(a+b).

8a−12b+5a+5b=13a−7b.

20. Re-Substitution Can Reject a Bad Expansion Quickly

If someone claims −3(2x−5)=−6x−15, test x=0. Original gives 15; claimed form gives −15. The claim is false.

Your Turn 1

1. Expand 6(x+4).
2. Expand −4(3x−2).
3. Simplify 2(x+5)+3x.
4. Simplify 5(2a−b)−3(a+2b).

Answers

6x+24. −12x+8. 5x+10. 10a−5b−3a−6b=7a−11b.

21. Double-Bracket Expansion Uses the Same Distributive Law Twice

In (x+3)(x+5), every term in the first bracket multiplies every term in the second.

22. Teacher Model 6: Two Positive Brackets

(x+3)(x+5)=x²+5x+3x+15=x²+8x+15.

23. Four Products Can Become Three Terms

The two middle products are often like terms and combine after expansion.

24. Teacher Model 7: One Negative Constant

(x+4)(x−6)=x²−6x+4x−24=x²−2x−24.

25. Teacher Model 8: Both Brackets Contain Negative Terms

(x−2)(x−7)=x²−7x−2x+14=x²−9x+14.

26. Teacher Model 9: Coefficient Before x

(2x+3)(x+4)=2x²+8x+3x+12=2x²+11x+12.

27. Keep the Four Products Visible Until Sign Control Is Secure

Skipping directly from brackets to a trinomial may be fast only after the learner can reliably account for every product.

28. An Area Model Explains Double Distribution

For positive dimensions, a rectangle with sides x+3 and x+5 can be partitioned into areas x², 5x, 3x and 15. The algebraic identity remains valid beyond the physical picture.

29. Expand a Product Before Combining It With Other Terms

(x+2)(x+3)+4x becomes x²+5x+6+4x=x²+9x+6.

30. Teacher Model 10: Difference of Two Products

(x+5)(x+1)−(x+2)(x−3).

First product: x²+6x+5.

Second product: x²−x−6.

Subtract the entire second expression:

x²+6x+5−x²+x+6=7x+11.

31. A Minus Before an Expanded Expression Reverses Every Term

This is one of the most common Secondary 2 algebra control points.

Your Turn 2

1. Expand (x+6)(x+2).
2. Expand (x−3)(x+8).
3. Expand (2x−5)(x+3).
4. Simplify (x+4)(x+1)−(x−2)(x+3).

Answers

x²+8x+12. x²+5x−24. 2x²+x−15. First x²+5x+4; second x²+x−6; difference=4x+10.

32. Equivalent Forms Make Different Information Visible

(x+3)(x+4) and x²+7x+12 are equivalent. The product form shows factors; the expanded form shows coefficients.

33. Choose a Form According to the Next Mathematical Job

Use expanded form when collecting terms or comparing coefficients. Use factor form when a common factor, zero product or structural pattern matters. Chapter 3 develops that reverse direction in depth.

34. A Formula Is a Compressed Relationship Between Quantities

C=3n+5 can be read as a fixed amount 5 plus 3 for each unit of n.

35. The Subject Is the Variable Standing Alone

In A=lw, A is the subject. In v=u+at, v is the subject.

36. Substitution Replaces a Symbol With Its Complete Value

If x=−4 in 3x+7, write 3(−4)+7, not 3−4+7.

37. Teacher Model 11: Direct Substitution

Given y=5x−2, find y when x=7.

y=5(7)−2=33.

38. Teacher Model 12: Two Variables

Given P=2a+3b, with a=4 and b=5:

P=2(4)+3(5)=8+15=23.

39. Negative Substitution Must Be Bracketed Before Powers Are Applied

If x=−2, then x²=(−2)²=4.

The written form −2² means −(2²)=−4 under standard operation order.

40. Teacher Model 13: Negative Value and Powers

Given y=x²+3x−4 and x=−2:

y=(−2)²+3(−2)−4=4−6−4=−6.

41. Formulae Can Carry Units

For A=½bh, if b and h are in centimetres, A is in square centimetres.

42. Teacher Model 14: Geometry Formula

A=½bh, b=12 cm, h=7 cm.

A=½×12×7=42 cm².

43. Formulae With Denominators May Have Restrictions

For q=10/(x+3), x=−3 is not permitted because the denominator would be zero.

44. Teacher Model 15: Substitution Into a Denominator

q=10/(x+3), x=2 gives q=10/5=2.

Your Turn 3

1. y=4x+3, x=5.
2. y=4x+3, x=−2.
3. P=2a+5b, a=3, b=4.
4. A=½bh, b=14 cm, h=9 cm.
5. q=12/(x−1), x=5. State the excluded input.

Answers

23. −5. 26. 63 cm². q=3 and x≠1.

45. Changing the Subject Preserves the Relationship

Changing the subject means rewriting the same formula so that another variable stands alone.

46. Use Inverse Operations, Not “Move and Change Sign”

From y=3x+5, subtract 5 from both sides and divide both sides by 3:

x=(y−5)/3.

47. Teacher Model 16: Multiplied Variable

A=lw. Make w the subject.

w=A/l, assuming l≠0.

48. Teacher Model 17: Formula With a Fraction

A=bh/2. Make h the subject.

2A=bh, so h=2A/b.

49. Teacher Model 18: Subject Inside a Sum

P=2l+2w. Make w the subject.

P−2l=2w, so w=(P−2l)/2.

50. Equivalent Rearranged Forms Can Look Different

(P−2l)/2 and P/2−l are equivalent. Choose the form that keeps the next calculation clear.

51. Protect the Whole Numerator

x=(y−5)/3 is not the same as x=y−5/3.

52. Bridge: Subject Inside a Bracket

Where this depth is part of the current school sequence, y=a(x+b) can be rearranged by dividing by a first, then subtracting b:

x=y/a−b, for a≠0.

53. Bridge: Variable in a Denominator

From v=d/t, multiply by t and divide by v:

t=d/v, where the relevant divisions are permitted.

54. Units Can Reject a Broken Rearrangement

If v=d/t, speed units are distance/time. A proposed v=dt would have units distance×time and cannot represent speed.

Your Turn 4

1. Make x the subject of y=5x−7.
2. Make w the subject of A=lw.
3. Make h the subject of A=bh/2.
4. Make l the subject of P=2l+2w.
5. Bridge: make x the subject of y=a(x+b).

Answers

x=(y+7)/5. w=A/l. h=2A/b. l=(P−2w)/2. x=y/a−b.

55. An Identity Is True for Every Permitted Value

2(x+3)=2x+6 is an identity. It is not asking for one special x-value.

56. An Equation May Be True for One, Several, All or No Values

• 2x+3=11 → one solution;
• x²=9 → two real solutions;
• 2(x+3)=2x+6 → all real x;
• 2x+3=2x+7 → no solution.

57. Squaring a Sum Produces a Middle Term Twice

(a+b)²=a²+2ab+b².

58. Teacher Model 19: Expand a Square

(x+4)²=(x+4)(x+4)=x²+4x+4x+16=x²+8x+16.

59. Squaring Each Term Separately Is Wrong

(x+4)² is not x²+16. The missing middle term is 8x.

60. Square of a Difference

(a−b)²=a²−2ab+b².

61. Teacher Model 20: Expand (2x−3)²

(2x−3)²=4x²−12x+9.

62. Opposite-Sign Brackets Produce a Difference of Squares

(a+b)(a−b)=a²−b².

63. Teacher Model 21: Difference of Squares Identity

(3x+5)(3x−5)=9x²−15x+15x−25=9x²−25.

64. The Middle Terms Cancel Because They Are Opposites

The cancellation is a consequence of expansion, not a pattern to use without conditions.

65. Recognising an Identity Requires All Parts to Match

4x²−12x+9=(2x−3)² because the end terms are squares and the middle term is −2(2x)(3).

66. Similar-Looking End Terms Are Not Enough

4x²−10x+9 is not (2x−3)² because the middle term would have to be −12x.

67. Identity Recognition Prepares the Reverse Direction

Chapter 3 will use these patterns in reverse during factorisation. This chapter establishes why the patterns are true.

Your Turn 5

1. Expand (x+5)².
2. Expand (x−6)².
3. Expand (2x+3)(2x−3).
4. Decide whether 9x²−24x+16 is a perfect square. Explain.

Answers

x²+10x+25. x²−12x+36. 4x²−9. Yes: (3x−4)² because −2(3x)(4)=−24x.

68. Identity Proof Uses Valid Algebraic Transformation

To prove (x+3)²=x²+6x+9, expand the left side until it becomes the right side.

69. Checking Three Numbers Does Not Prove an Identity

A false equation can agree for a few chosen values. Numerical testing can reject a false statement but cannot replace an algebraic proof of a general identity.

70. Counterexamples Can Disprove a False Identity Quickly

Claim: (x+2)²=x²+4. At x=1, left side=9 and right side=5. One counterexample is enough to disprove the universal claim.

71. Verification Method 1: Re-Substitute a Convenient Value

If two expressions are supposed to be equivalent, a mismatch at one legal value proves they are not equivalent.

72. Verification Method 2: Reverse the Transformation

After changing the subject, substitute one consistent set of values into both forms. After factorisation later, expand back.

73. Verification Method 3: Check Units

In d=vt, km/h×h gives km. If a rearrangement produces incompatible units, something has broken.

74. Teacher Model 22: Verify a Rearranged Formula

Original C=3n+5. Rearranged n=(C−5)/3.

Choose n=4. Original gives C=17. Rearranged gives (17−5)/3=4. The two forms agree.

75. Teacher Model 23: Verify an Expansion

Claim: (2x−3)(x+5)=2x²+7x−15.

At x=1, original=(−1)(6)=−6. Claimed expansion=2+7−15=−6. The numerical check is consistent; full expansion confirms it for all x.

76. A Good Check Uses a Different Route

Repeating the same expansion twice can repeat the same sign error. Use substitution, structure or reverse operation when possible.

77. Misconception Clinic: Multiply Only the First Term in a Bracket

Repair: the external factor multiplies every term inside.

78. Misconception Clinic: Negative Times Negative Stays Negative

Repair: signed-number multiplication still controls algebra.

79. Misconception Clinic: Combine Unlike Terms

3x+4x² cannot become 7x³ or 7x². The terms are not alike.

80. Misconception Clinic: Square a Sum by Squaring Each Part

(a+b)² includes the middle term 2ab.

81. Misconception Clinic: Forget the Minus Before the Second Expanded Product

When subtracting a whole expression, every term of that expression changes sign.

82. Misconception Clinic: Substitute −3 Without Brackets Into x²

Write (−3)², not −3², when the variable’s value is −3.

83. Misconception Clinic: Change the Subject by Visual Transposition

Repair: state and apply inverse operations to both sides.

84. Misconception Clinic: Divide Only One Term of a Numerator

(y−5)/3 means the entire difference is divided by 3.

85. Misconception Clinic: One Numerical Check Proves an Identity

Repair: numerical checks test examples; algebra proves the universal statement.

86. Misconception Clinic: Correct Final Answer Means Every Line Was Valid

A later arithmetic accident can sometimes repair an earlier error. Check the first line where equivalence stops being true.

87. Guided Practice A: Single Brackets

1. Expand 7(x+2).
2. Expand −5(2x−3).
3. Simplify 4(x+1)+3x.
4. Simplify 3(2a−b)−2(a+4b).

Solutions

7x+14. −10x+15. 7x+4. 6a−3b−2a−8b=4a−11b.

88. Guided Practice B: Double Brackets

1. (x+2)(x+9).
2. (x−5)(x+4).
3. (3x+2)(x−6).
4. (2x−1)(2x+5).

Solutions

x²+11x+18. x²−x−20. 3x²−16x−12. 4x²+8x−5.

89. Guided Practice C: Substitution

1. Evaluate 3x²−2x+1 at x=−2.
2. Evaluate 2a−3b for a=−1, b=4.
3. Given A=πr², find A when r=3 using π≈3.142.

Solutions

12+4+1=17. −2−12=−14. A≈28.278 square units.

90. Guided Practice D: Formulae

1. Make x the subject of y=4x+9.
2. Make b the subject of A=bh/2.
3. Make w the subject of P=2l+2w.
4. Bridge: make t the subject of v=d/t.

Solutions

x=(y−9)/4. b=2A/h. w=(P−2l)/2. t=d/v.

91. Guided Practice E: Identities

1. Expand (x+7)².
2. Expand (2x−5)².
3. Expand (4x+1)(4x−1).
4. Is x²+12x+36 a perfect square? State the form.

Solutions

x²+14x+49. 4x²−20x+25. 16x²−1. Yes, (x+6)².

92. Guided Practice F: Error Analysis

A student writes:

−2(3x−4)=−6x−8.

Identify the first false product and correct the line.

Answer

−2×−4=+8, so the correct expansion is −6x+8.

93. Guided Practice G: Mixed Job Selection

1. Expand 4(x−3).
2. Evaluate 4(x−3) at x=8.
3. Decide whether 4(x−3)=4x−12 is an identity.
4. Solve 4(x−3)=20.

Answers

4x−12. 20. Yes. x=8.

94. Challenge Practice: Two Valid Forms, One Choice

E=(x+3)(x+5)=x²+8x+15.

1. Which form is faster for evaluating E at x=−3?
2. Which form is clearer for identifying the coefficient of x?
3. Explain why neither form is “more correct”.

Answer

Factor form gives zero immediately. Expanded form shows coefficient 8. Both are equivalent; usefulness depends on the task.

95. Challenge Practice: Verify a False Identity

Claim: (2x+1)²=4x²+1.

Worked response

At x=1, left=9 and right=5, so the claim is false. Correct expansion: (2x+1)²=4x²+4x+1.

96. Challenge Practice: Formula and Units

Using d=vt, a speed is 72 km/h and time is 25 minutes. Find d.

Worked solution

25 min=25/60 h. d=72×25/60=30 km. The conversion is required before substitution because speed is measured per hour.

97. Assessment Method: Identify the Mathematical Job

• expand → distribute;
• simplify → expand if needed, then collect like terms;
• evaluate → substitute values;
• change subject → isolate a variable by inverse operations;
• verify identity → transform algebraically or disprove with a counterexample;
• compare forms → ask which information each form reveals.

98. Assessment Method: Preserve Every Sign

Write signed terms explicitly. Most bracket mistakes are not mysterious algebra errors; they are lost multiplication signs or lost negative values.

99. Assessment Method: Keep the Four Products Visible

During double-bracket expansion, display all products until control is secure.

100. Assessment Method: Do Not Collect Before the Terms Exist

Expand first. Then collect like terms. Combining across an unexpanded bracket often hides the operation structure.

101. Assessment Method: Protect Complete Fractions

Use brackets around numerators and denominators when several terms belong together.

102. Assessment Method: Verify With a Different Route

• expansion → substitute a value;
• formula rearrangement → substitute a consistent example into both forms;
• identity → expand one side to the other;
• units → check dimensional consistency.

103. Oral Classroom Check

1. What is the difference between a term and a factor?
2. Why does 3(x+5) require two products?
3. Why does −2(x−4) end with +8?
4. Why can 3x and 5x be collected but 3x and 5x² cannot?
5. Why is (x+3)² not x²+9?
6. What makes two expressions equivalent?
7. What is the subject of a formula?
8. Why should negative substituted values be bracketed?
9. What is an identity?
10. Why does one counterexample disprove an identity claim?

104. Exit Ticket

1. Expand −3(2x−7).
2. Simplify 4(x+2)−3(x−5).
3. Expand (x+4)(x−6).
4. Expand (2x−3)².
5. Expand (5x+2)(5x−2).
6. Evaluate x²−4x+1 at x=−2.
7. Make x the subject of y=6x+11.
8. Make h the subject of A=bh/2.
9. Explain why (x+5)²=x²+25 is false.
10. State one way to verify a formula rearrangement.

Exit-ticket solutions

−6x+21. 4x+8−3x+15=x+23. x²−2x−24. 4x²−12x+9. 25x²−4. 4+8+1=13. x=(y−11)/6. h=2A/b. The correct square includes 10x: x²+10x+25; x=1 gives 36 versus 26 for the false claim. Substitute one consistent set of values into the original and rearranged formula and confirm both agree.

105. Homework: Retrieval, Variation and Transfer

Layer 1 — Retrieval

• state the distributive law;
• define term, factor, expression, formula, subject and identity;
• state the three standard identities used in this classroom;
• explain why negative substitution needs brackets;
• explain why units can detect a broken formula.

Layer 2 — Variation

• four single-bracket expansions;
• four double-bracket expansions;
• three substitutions including negative values;
• three formula rearrangements;
• three identity-recognition questions;
• one error-analysis question.

Layer 3 — Transfer

Create one expression that can be written in both product and expanded form. Explain one situation in which each form would be more useful. Then create a false identity that passes one easy numerical test but fails another.

106. The Seven-Day Return Cycle

1. Day 0: single brackets, double brackets and formula substitution.
2. Day 1: one negative-expansion task, one double-bracket task and one change-of-subject task.
3. Day 3: four unlabeled mixed algebra jobs and one identity check.
4. Day 7: complete a changed exit ticket without notes and explain the verification method for two answers.

107. A 60-Minute Teaching Lesson

1. 5 minutes: Secondary 1 algebra retrieval.
2. 15 minutes: single brackets and sign control.
3. 15 minutes: double brackets.
4. 10 minutes: substitution and formula meaning.
5. 10 minutes: basic changing of subject.
6. 5 minutes: exit ticket.

108. A 90-Minute Teaching Lesson

1. 10 minutes: structure diagnostic.
2. 20 minutes: single and double brackets.
3. 15 minutes: equivalent forms and verification.
4. 15 minutes: substitution and formulae.
5. 15 minutes: changing the subject.
6. 10 minutes: identities and counterexamples.
7. 5 minutes: exit ticket and return date.

109. The Full Expansion Routine

read signs → identify every term in each bracket → form every required product → simplify products → collect like terms → substitute one value to check if useful.

110. The Full Formula Routine

name quantities → identify the current subject → substitute complete values or isolate the target variable → preserve brackets and units → verify with a consistent example.

111. Connect Back to Secondary 1

This classroom grows directly from Secondary 1 Chapter 2: Directed Numbers, Chapter 6: Algebraic Language, Expressions, Formulae and nth-Term Patterns, and Chapter 7: Linear Equations and Problem Formation.

112. Specialist Companions

• Secondary 2 Mathematics Learning Guide | Algebraic Factorisation and Structural Control
• Secondary 2 Mathematics Learning Guide | Formulae, Change of Subject and Substitution Control

113. Why This Chapter Matters

Expansion and identities are the forward direction of structures that Chapter 3 will learn to recognise and reverse. Formula control will also reappear in equations, graphs, geometry, mensuration and real-world modelling.

114. Ready for Chapter 3?

You are ready to continue when you can do all of the following without prompts:

• distinguish terms from factors;
• expand a single bracket with positive or negative multiplier;
• expand two brackets and collect the middle terms;
• subtract a whole expanded expression without losing signs;
• substitute negative values using brackets;
• evaluate formulae with correct units;
• make a variable the subject in common lower-secondary forms;
• recognise and expand the square-of-sum, square-of-difference and difference-of-squares identities;
• distinguish an identity from an equation with a particular solution;
• use a counterexample to disprove a false universal claim;
• verify equivalent forms through substitution or reverse transformation.

If one item is weak, return to the smallest section that owns it and solve a changed example. If all are stable, continue to Chapter 3: Factorisation and Algebraic Fractions, where the expansion structures learned here are read in reverse and restrictions become part of the algebra.