Sec 3 G3 Maths Tutor Bukit Timah | Algebraic Identities

Secondary 3 G3 Mathematics Tutor Bukit Timah | Algebraic Identities and Factorisation by Grouping — Did you know? Some of the hardest-looking algebra questions become much simpler when you can recognise an expression in two different forms. A bracketed product and an expanded polynomial may be the same mathematical object. The practical skill is knowing when to expand, when to factorise and how to check that neither operation has changed the original expression.

Quick answer: which identity should I use?

(a+b)² = a²+2ab+b². The middle cross-product matters.
(a−b)² = a²−2ab+b². The last square remains positive.
a²−b² = (a−b)(a+b). The cross-products cancel when expanded.
Look for a common factor before trying a more complicated identity.
In grouping, arrange terms so the same complete bracket appears in both groups.
After factorisation, expand the answer to verify every sign and coefficient.

These skills sit within the 2027 Singapore–Cambridge SEC G3 Mathematics (K310) algebraic expressions and formulae content. The syllabus includes expansion, special algebraic identities, factorisation and grouping structures. A school’s chapter sequence can differ, so this resource should support — not replace — the student’s current class notes and school instructions.

Sections: Understand identities · Worked examples · Error checks · Practice plan · FAQs.

Why algebraic identities work

A bracket is a structure, not punctuation

When we write (x+3)(x+5), every term in the first bracket multiplies every term in the second. The four products combine into x²+8x+15. The brackets are not merely decorative; they express how numbers and variables are related through multiplication.

Learning expansion is not only about moving symbols. It gives a reliable way to test any later factorisation. If a student claims x²+8x+15=(x+2)(x+6), expansion immediately exposes the difference: that proposed product gives x²+8x+12.

Why the square of a sum needs a middle term

To square a bracket, multiply it by itself. (a+b)(a+b) produces a², ab, ba and b². The two identical cross-products combine into 2ab. That is why (a+b)² does not simplify to a²+b².

A quick numerical example makes the issue visible: (2+3)²=25, whereas 2²+3²=13. The missing twelve equals 2×2×3. This is a useful quick check, but the algebraic expansion explains why the rule holds for all values.

Minus signs should be read, not guessed

In (a−b)², both mixed products are negative, but (−b)(−b)=b² is positive. The result is a²−2ab+b². In contrast, a²−b² is a difference of squares, not a squared difference, and factors into (a−b)(a+b).

These forms look similar when written quickly, which is why sign mistakes are common. A tutor can ask students to write the two brackets explicitly and expand them before memorising. Once the cross-products are understood, the signs become consequences rather than unpredictable decorations.

Factorisation is the reverse trip

Expansion changes a product into a sum; factorisation changes a suitable sum into a product. A correct factorisation, when expanded, must reproduce the exact original polynomial. The same equivalence makes factorisation useful in equation solving, simplification and graph analysis.

Students may recognise a pair of numbers whose sum looks correct but forget to check their product. A stronger routine is to multiply the proposed factors to check the leading coefficient, middle term and constant. This reverse-trip habit repairs many plausible but incorrect answers.

Grouping requires a repeated complete factor

An expression such as ax+bx+kay+kby can first be organised as (a+b)x+k(a+b)y. Now the whole bracket a+b repeats, so we can factor it to obtain (a+b)(x+ky). A clear group-by-group explanation makes the method auditable.

Grouping does not give permission to cancel pieces of addition or invent a common factor that is not present. Every bracket has to be a complete multiplicative factor. With negative coefficients, it may be useful to factor out a negative value so the two repeated brackets match exactly.

A short sequence for choosing the method

First, look for a common factor shared by every term. Second, ask whether the expression matches a special identity: a perfect-square trinomial or a two-square difference. Third, consider grouping, ordinary binomial expansion or a suitable quadratic method.

This priority order reduces unnecessary work. A pupil who immediately searches for two factors in 6x²+12x may miss that the easy answer is simply 6x(x+2). In Mathematics, noticing a simple structure is often more powerful than applying a more elaborate technique.

Understanding the identity makes transfer easier

One school question might ask a student to expand; the next might ask them to factorise a rearranged expression. A student who understands that the two processes are inverses can adapt to either direction without learning a separate list of disconnected tricks.

Good practice therefore alternates the directions. Expand a bracketed square, then factorise the result; simplify by common factors, then expand to verify. The answer should be expressed clearly, and any eventual equation solving should distinguish an identity from a condition true only at particular x values.

Fifty worked examples, each with a check

These are constructed teaching examples, not reproduced or predicted examination questions. Each group gives five numerical variations of one underlying algebraic move. First attempt a question with the worked line hidden; then check it by expansion or substitution. The objective is to become more reliable at selecting a valid identity, not to finish every item in one sitting.

A squared sum

This group focuses on a squared sum. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 1. Question: Expand (x+2)². Method: (x+2)² means (x+2)(x+2), not two separate squares. Multiply all four terms, giving x²+4x+4. Check: Set x=1: both forms equal 3² = 9. Watch for: The middle term 2×x×2 is essential and cannot disappear.

Try it differently: Change the plus sign to minus; predict the new middle term. Before expanding, predict which powers of x can appear. The structure of the factors often tells you this even before calculation. Construct a new example with different coefficients and test it at x=1.

A squared difference

This group focuses on a squared difference. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 2. Question: Expand (x−3)² completely. Method: Write (x−3)(x−3). The two cross-products are each −3x and the constant square is positive, so x²−6x+9. Check: At x=0, the squared bracket and the expanded polynomial both equal 9. Watch for: The constant is positive even though the original bracket contains a negative term.

Try it differently: Explain why (x−3)² differs from x²−9. Return to the factorised or expanded form in the checking line; a valid result must match every coefficient. Use a numerical counterexample to explain why one tempting identity is invalid.

The difference of two squares

This group focuses on the difference of two squares. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 3. Question: Factorise x²−16. Method: Both terms are squares. Using A²−B²=(A−B)(A+B), the expression becomes (x−4)(x+4). Check: Multiplication gives x²+4x−4x−16 = x²−16. Watch for: An expression x²−16 is not the same as (x−4)².

Try it differently: Find its zeros if the factorised expression equals zero. Explain why the middle term has the sign shown rather than appealing only to a memorised chant. Turn an expansion question into a factorisation question using the computed polynomial.

Grouping four terms

This group focuses on grouping four terms. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 4. Question: Factorise 2x+3x+4y+6y. Method: Group the x and y terms: (2+3)x+(4+6)y = 5x+10y. Extract the common factor 5: 5(x+2y). Check: Expanding the factorised form gives the same combined coefficients 5 for x and 10 for y. Watch for: Grouping works because the same complete factor is shared; it is not arbitrary rearrangement.

Try it differently: Regroup as 2(x+2y)+3(x+2y) and verify the same final product. Hide the method and reproduce one legal line independently. Being able to start unaided matters more than recognising the printed answer. Explain why the same expression may have several equivalent-looking but differently ordered factor products.

A repeated bracket from grouping

This group focuses on a repeated bracket from grouping. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 5. Question: Factorise 2x+2y+4x+4y. Method: The terms can be grouped as 2(x+y)+4(x+y). Extract the repeated (x+y) bracket to obtain (2+4)(x+y)=6(x+y). Check: Expanding gives 6x+6y, equal to the original collected like terms. Watch for: Do not cancel or drop one repeated bracket; it appears as a common factor.

Try it differently: Evaluate both forms at x=1 and y=2 to confirm the relationship. Try the same idea with a negative coefficient and notice where brackets are essential. Substitute x=0 and another value into both forms to catch a coefficient or constant mistake.

A greatest obvious common factor

This group focuses on a greatest obvious common factor. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 6. Question: Factorise 3x²+6x. Method: Each term contains 3x. Factor it out to leave 3x(x+2). This is an algebraic product, not a sum of isolated cancelled terms. Check: Expanding 3x(x+2) gives 3x²+6x, the original expression. Watch for: Taking out x without the numerical coefficient is correct but may leave a form that is not completely factored.

Try it differently: Use the two factors to solve 3x²+6x=0. Ask whether a common factor could be removed before applying the more complicated identity. Ask whether any remaining factor can be extracted to make the answer more completely factorised.

Expand two different simple brackets

This group focuses on expand two different simple brackets. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 7. Question: Expand (x+2)(x+5). Method: The four products are x², 2x, 5x and 10; combining like terms gives x²+7x+10. Check: At x=0 both forms equal 10, so at least the constant term is checked independently. Watch for: The coefficient of x is the sum 7, not the product 10.

Try it differently: Reverse this result: factorise the expanded polynomial without looking at the original brackets. Write the first likely sign error in a small correction log and name the operation that prevents it. Construct a new example with different coefficients and test it at x=1.

Recognise a perfect square in reverse

This group focuses on recognise a perfect square in reverse. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 8. Question: Factorise x²+4x+4. Method: The first and last terms are x² and 2². The middle term is twice the product x×2, so the expression is (x+2)². Check: Expanding the answer restores all three original terms in the same order. Watch for: Two squared end terms alone are not enough; verify the middle coefficient exactly.

Try it differently: Write a corresponding expression for (x−2)² and explain its sign change. Compare this pattern with the previous one: which term or coefficient tells you the method should change? Use a numerical counterexample to explain why one tempting identity is invalid.

Non-unit squares in a difference

This group focuses on non-unit squares in a difference. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 9. Question: Factorise 4x²−25. Method: The two squares are (2x)² and 5². Their difference factors as (2x−5)(2x+5). Check: On expansion, the opposite cross-products −10x and +10x cancel. Watch for: The factors involve 2x, because the coefficient of x² is 4, not 1.

Try it differently: Solve 4x²−25=0 and check both roots. Return after two days without a worksheet heading; decide whether to expand, factorise or group based on the expression. Turn an expansion question into a factorisation question using the computed polynomial.

Full distribution with non-unit coefficients

This group focuses on full distribution with non-unit coefficients. Pay attention to whole brackets and complete products. When the next group begins, identify how its mathematical structure differs; an assessment often removes the helpful chapter label and leaves you to make that choice yourself.

Worked example 10. Question: Expand (2x+2)(3x+4). Method: Multiply all four pairs: 6x² + 8x + 6x + 8. Combine the middle terms to obtain 6x²+14x+8. Check: At x=0 the product form and expanded form both equal 8, and the leading x² coefficient is 2×3=6. Watch for: Failing to include both cross-products changes the middle coefficient.

Try it differently: Check the result by expanding with the two brackets written in reversed order. Give the solution a final reverse check using expansion, not only the fact that it looks neat. Explain why the same expression may have several equivalent-looking but differently ordered factor products.

A squared sum

Worked example 11. Question: Expand (x+3)². Method: (x+3)² means (x+3)(x+3), not two separate squares. Multiply all four terms, giving x²+6x+9. Check: Set x=1: both forms equal 4² = 16. Watch for: The middle term 2×x×3 is essential and cannot disappear.

Try it differently: Change the plus sign to minus; predict the new middle term. Before expanding, predict which powers of x can appear. The structure of the factors often tells you this even before calculation. Substitute x=0 and another value into both forms to catch a coefficient or constant mistake.

A squared difference

Worked example 12. Question: Expand (x−4)² completely. Method: Write (x−4)(x−4). The two cross-products are each −4x and the constant square is positive, so x²−8x+16. Check: At x=0, the squared bracket and the expanded polynomial both equal 16. Watch for: The constant is positive even though the original bracket contains a negative term.

Try it differently: Explain why (x−4)² differs from x²−16. Return to the factorised or expanded form in the checking line; a valid result must match every coefficient. Ask whether any remaining factor can be extracted to make the answer more completely factorised.

The difference of two squares

Worked example 13. Question: Factorise x²−25. Method: Both terms are squares. Using A²−B²=(A−B)(A+B), the expression becomes (x−5)(x+5). Check: Multiplication gives x²+5x−5x−25 = x²−25. Watch for: An expression x²−25 is not the same as (x−5)².

Try it differently: Find its zeros if the factorised expression equals zero. Explain why the middle term has the sign shown rather than appealing only to a memorised chant. Construct a new example with different coefficients and test it at x=1.

Grouping four terms

Worked example 14. Question: Factorise 3x+4x+9y+12y. Method: Group the x and y terms: (3+4)x+(9+12)y = 7x+21y. Extract the common factor 7: 7(x+3y). Check: Expanding the factorised form gives the same combined coefficients 7 for x and 21 for y. Watch for: Grouping works because the same complete factor is shared; it is not arbitrary rearrangement.

Try it differently: Regroup as 3(x+3y)+4(x+3y) and verify the same final product. Hide the method and reproduce one legal line independently. Being able to start unaided matters more than recognising the printed answer. Use a numerical counterexample to explain why one tempting identity is invalid.

A repeated bracket from grouping

Worked example 15. Question: Factorise 3x+3y+5x+5y. Method: The terms can be grouped as 3(x+y)+5(x+y). Extract the repeated (x+y) bracket to obtain (3+5)(x+y)=8(x+y). Check: Expanding gives 8x+8y, equal to the original collected like terms. Watch for: Do not cancel or drop one repeated bracket; it appears as a common factor.

Try it differently: Evaluate both forms at x=1 and y=2 to confirm the relationship. Try the same idea with a negative coefficient and notice where brackets are essential. Turn an expansion question into a factorisation question using the computed polynomial.

A greatest obvious common factor

Worked example 16. Question: Factorise 4x²+12x. Method: Each term contains 4x. Factor it out to leave 4x(x+3). This is an algebraic product, not a sum of isolated cancelled terms. Check: Expanding 4x(x+3) gives 4x²+12x, the original expression. Watch for: Taking out x without the numerical coefficient is correct but may leave a form that is not completely factored.

Try it differently: Use the two factors to solve 4x²+12x=0. Ask whether a common factor could be removed before applying the more complicated identity. Explain why the same expression may have several equivalent-looking but differently ordered factor products.

Expand two different simple brackets

Worked example 17. Question: Expand (x+3)(x+6). Method: The four products are x², 3x, 6x and 18; combining like terms gives x²+9x+18. Check: At x=0 both forms equal 18, so at least the constant term is checked independently. Watch for: The coefficient of x is the sum 9, not the product 18.

Try it differently: Reverse this result: factorise the expanded polynomial without looking at the original brackets. Write the first likely sign error in a small correction log and name the operation that prevents it. Substitute x=0 and another value into both forms to catch a coefficient or constant mistake.

Recognise a perfect square in reverse

Worked example 18. Question: Factorise x²+6x+9. Method: The first and last terms are x² and 3². The middle term is twice the product x×3, so the expression is (x+3)². Check: Expanding the answer restores all three original terms in the same order. Watch for: Two squared end terms alone are not enough; verify the middle coefficient exactly.

Try it differently: Write a corresponding expression for (x−3)² and explain its sign change. Compare this pattern with the previous one: which term or coefficient tells you the method should change? Ask whether any remaining factor can be extracted to make the answer more completely factorised.

Non-unit squares in a difference

Worked example 19. Question: Factorise 9x²−36. Method: The two squares are (3x)² and 6². Their difference factors as (3x−6)(3x+6). Check: On expansion, the opposite cross-products −18x and +18x cancel. Watch for: The factors involve 3x, because the coefficient of x² is 9, not 1.

Try it differently: Solve 9x²−36=0 and check both roots. Return after two days without a worksheet heading; decide whether to expand, factorise or group based on the expression. Construct a new example with different coefficients and test it at x=1.

Full distribution with non-unit coefficients

Worked example 20. Question: Expand (3x+3)(4x+5). Method: Multiply all four pairs: 12x² + 15x + 12x + 15. Combine the middle terms to obtain 12x²+27x+15. Check: At x=0 the product form and expanded form both equal 15, and the leading x² coefficient is 3×4=12. Watch for: Failing to include both cross-products changes the middle coefficient.

Try it differently: Check the result by expanding with the two brackets written in reversed order. Give the solution a final reverse check using expansion, not only the fact that it looks neat. Use a numerical counterexample to explain why one tempting identity is invalid.

A squared sum

Worked example 21. Question: Expand (x+4)². Method: (x+4)² means (x+4)(x+4), not two separate squares. Multiply all four terms, giving x²+8x+16. Check: Set x=1: both forms equal 5² = 25. Watch for: The middle term 2×x×4 is essential and cannot disappear.

Try it differently: Change the plus sign to minus; predict the new middle term. Before expanding, predict which powers of x can appear. The structure of the factors often tells you this even before calculation. Turn an expansion question into a factorisation question using the computed polynomial.

A squared difference

Worked example 22. Question: Expand (x−5)² completely. Method: Write (x−5)(x−5). The two cross-products are each −5x and the constant square is positive, so x²−10x+25. Check: At x=0, the squared bracket and the expanded polynomial both equal 25. Watch for: The constant is positive even though the original bracket contains a negative term.

Try it differently: Explain why (x−5)² differs from x²−25. Return to the factorised or expanded form in the checking line; a valid result must match every coefficient. Explain why the same expression may have several equivalent-looking but differently ordered factor products.

The difference of two squares

Worked example 23. Question: Factorise x²−36. Method: Both terms are squares. Using A²−B²=(A−B)(A+B), the expression becomes (x−6)(x+6). Check: Multiplication gives x²+6x−6x−36 = x²−36. Watch for: An expression x²−36 is not the same as (x−6)².

Try it differently: Find its zeros if the factorised expression equals zero. Explain why the middle term has the sign shown rather than appealing only to a memorised chant. Substitute x=0 and another value into both forms to catch a coefficient or constant mistake.

Grouping four terms

Worked example 24. Question: Factorise 4x+5x+16y+20y. Method: Group the x and y terms: (4+5)x+(16+20)y = 9x+36y. Extract the common factor 9: 9(x+4y). Check: Expanding the factorised form gives the same combined coefficients 9 for x and 36 for y. Watch for: Grouping works because the same complete factor is shared; it is not arbitrary rearrangement.

Try it differently: Regroup as 4(x+4y)+5(x+4y) and verify the same final product. Hide the method and reproduce one legal line independently. Being able to start unaided matters more than recognising the printed answer. Ask whether any remaining factor can be extracted to make the answer more completely factorised.

A repeated bracket from grouping

Worked example 25. Question: Factorise 4x+4y+6x+6y. Method: The terms can be grouped as 4(x+y)+6(x+y). Extract the repeated (x+y) bracket to obtain (4+6)(x+y)=10(x+y). Check: Expanding gives 10x+10y, equal to the original collected like terms. Watch for: Do not cancel or drop one repeated bracket; it appears as a common factor.

Try it differently: Evaluate both forms at x=1 and y=2 to confirm the relationship. Try the same idea with a negative coefficient and notice where brackets are essential. Construct a new example with different coefficients and test it at x=1.

A greatest obvious common factor

Worked example 26. Question: Factorise 5x²+20x. Method: Each term contains 5x. Factor it out to leave 5x(x+4). This is an algebraic product, not a sum of isolated cancelled terms. Check: Expanding 5x(x+4) gives 5x²+20x, the original expression. Watch for: Taking out x without the numerical coefficient is correct but may leave a form that is not completely factored.

Try it differently: Use the two factors to solve 5x²+20x=0. Ask whether a common factor could be removed before applying the more complicated identity. Use a numerical counterexample to explain why one tempting identity is invalid.

Expand two different simple brackets

Worked example 27. Question: Expand (x+4)(x+7). Method: The four products are x², 4x, 7x and 28; combining like terms gives x²+11x+28. Check: At x=0 both forms equal 28, so at least the constant term is checked independently. Watch for: The coefficient of x is the sum 11, not the product 28.

Try it differently: Reverse this result: factorise the expanded polynomial without looking at the original brackets. Write the first likely sign error in a small correction log and name the operation that prevents it. Turn an expansion question into a factorisation question using the computed polynomial.

Recognise a perfect square in reverse

Worked example 28. Question: Factorise x²+8x+16. Method: The first and last terms are x² and 4². The middle term is twice the product x×4, so the expression is (x+4)². Check: Expanding the answer restores all three original terms in the same order. Watch for: Two squared end terms alone are not enough; verify the middle coefficient exactly.

Try it differently: Write a corresponding expression for (x−4)² and explain its sign change. Compare this pattern with the previous one: which term or coefficient tells you the method should change? Explain why the same expression may have several equivalent-looking but differently ordered factor products.

Non-unit squares in a difference

Worked example 29. Question: Factorise 16x²−49. Method: The two squares are (4x)² and 7². Their difference factors as (4x−7)(4x+7). Check: On expansion, the opposite cross-products −28x and +28x cancel. Watch for: The factors involve 4x, because the coefficient of x² is 16, not 1.

Try it differently: Solve 16x²−49=0 and check both roots. Return after two days without a worksheet heading; decide whether to expand, factorise or group based on the expression. Substitute x=0 and another value into both forms to catch a coefficient or constant mistake.

Full distribution with non-unit coefficients

Worked example 30. Question: Expand (4x+4)(5x+6). Method: Multiply all four pairs: 20x² + 24x + 20x + 24. Combine the middle terms to obtain 20x²+44x+24. Check: At x=0 the product form and expanded form both equal 24, and the leading x² coefficient is 4×5=20. Watch for: Failing to include both cross-products changes the middle coefficient.

Try it differently: Check the result by expanding with the two brackets written in reversed order. Give the solution a final reverse check using expansion, not only the fact that it looks neat. Ask whether any remaining factor can be extracted to make the answer more completely factorised.

A squared sum

Worked example 31. Question: Expand (x+5)². Method: (x+5)² means (x+5)(x+5), not two separate squares. Multiply all four terms, giving x²+10x+25. Check: Set x=1: both forms equal 6² = 36. Watch for: The middle term 2×x×5 is essential and cannot disappear.

A squared difference

Worked example 32. Question: Expand (x−6)² completely. Method: Write (x−6)(x−6). The two cross-products are each −6x and the constant square is positive, so x²−12x+36. Check: At x=0, the squared bracket and the expanded polynomial both equal 36. Watch for: The constant is positive even though the original bracket contains a negative term.

Try it differently: Explain why (x−6)² differs from x²−36. Return to the factorised or expanded form in the checking line; a valid result must match every coefficient. Use a numerical counterexample to explain why one tempting identity is invalid.

The difference of two squares

Worked example 33. Question: Factorise x²−49. Method: Both terms are squares. Using A²−B²=(A−B)(A+B), the expression becomes (x−7)(x+7). Check: Multiplication gives x²+7x−7x−49 = x²−49. Watch for: An expression x²−49 is not the same as (x−7)².

Grouping four terms

Worked example 34. Question: Factorise 5x+6x+25y+30y. Method: Group the x and y terms: (5+6)x+(25+30)y = 11x+55y. Extract the common factor 11: 11(x+5y). Check: Expanding the factorised form gives the same combined coefficients 11 for x and 55 for y. Watch for: Grouping works because the same complete factor is shared; it is not arbitrary rearrangement.

Try it differently: Regroup as 5(x+5y)+6(x+5y) and verify the same final product. Hide the method and reproduce one legal line independently. Being able to start unaided matters more than recognising the printed answer. Explain why the same expression may have several equivalent-looking but differently ordered factor products.

A repeated bracket from grouping

Worked example 35. Question: Factorise 5x+5y+7x+7y. Method: The terms can be grouped as 5(x+y)+7(x+y). Extract the repeated (x+y) bracket to obtain (5+7)(x+y)=12(x+y). Check: Expanding gives 12x+12y, equal to the original collected like terms. Watch for: Do not cancel or drop one repeated bracket; it appears as a common factor.

A greatest obvious common factor

Worked example 36. Question: Factorise 6x²+30x. Method: Each term contains 6x. Factor it out to leave 6x(x+5). This is an algebraic product, not a sum of isolated cancelled terms. Check: Expanding 6x(x+5) gives 6x²+30x, the original expression. Watch for: Taking out x without the numerical coefficient is correct but may leave a form that is not completely factored.

Try it differently: Use the two factors to solve 6x²+30x=0. Ask whether a common factor could be removed before applying the more complicated identity. Ask whether any remaining factor can be extracted to make the answer more completely factorised.

Expand two different simple brackets

Worked example 37. Question: Expand (x+5)(x+8). Method: The four products are x², 5x, 8x and 40; combining like terms gives x²+13x+40. Check: At x=0 both forms equal 40, so at least the constant term is checked independently. Watch for: The coefficient of x is the sum 13, not the product 40.

Recognise a perfect square in reverse

Worked example 38. Question: Factorise x²+10x+25. Method: The first and last terms are x² and 5². The middle term is twice the product x×5, so the expression is (x+5)². Check: Expanding the answer restores all three original terms in the same order. Watch for: Two squared end terms alone are not enough; verify the middle coefficient exactly.

Try it differently: Write a corresponding expression for (x−5)² and explain its sign change. Compare this pattern with the previous one: which term or coefficient tells you the method should change? Use a numerical counterexample to explain why one tempting identity is invalid.

Non-unit squares in a difference

Worked example 39. Question: Factorise 25x²−64. Method: The two squares are (5x)² and 8². Their difference factors as (5x−8)(5x+8). Check: On expansion, the opposite cross-products −40x and +40x cancel. Watch for: The factors involve 5x, because the coefficient of x² is 25, not 1.

Try it differently: Solve 25x²−64=0 and check both roots. Return after two days without a worksheet heading; decide whether to expand, factorise or group based on the expression. Turn an expansion question into a factorisation question using the computed polynomial.

Full distribution with non-unit coefficients

Worked example 40. Question: Expand (5x+5)(6x+7). Method: Multiply all four pairs: 30x² + 35x + 30x + 35. Combine the middle terms to obtain 30x²+65x+35. Check: At x=0 the product form and expanded form both equal 35, and the leading x² coefficient is 5×6=30. Watch for: Failing to include both cross-products changes the middle coefficient.

A squared sum

Worked example 41. Question: Expand (x+6)². Method: (x+6)² means (x+6)(x+6), not two separate squares. Multiply all four terms, giving x²+12x+36. Check: Set x=1: both forms equal 7² = 49. Watch for: The middle term 2×x×6 is essential and cannot disappear.

Try it differently: Change the plus sign to minus; predict the new middle term. Before expanding, predict which powers of x can appear. The structure of the factors often tells you this even before calculation. Substitute x=0 and another value into both forms to catch a coefficient or constant mistake.

A squared difference

Worked example 42. Question: Expand (x−7)² completely. Method: Write (x−7)(x−7). The two cross-products are each −7x and the constant square is positive, so x²−14x+49. Check: At x=0, the squared bracket and the expanded polynomial both equal 49. Watch for: The constant is positive even though the original bracket contains a negative term.

Try it differently: Explain why (x−7)² differs from x²−49. Return to the factorised or expanded form in the checking line; a valid result must match every coefficient. Ask whether any remaining factor can be extracted to make the answer more completely factorised.

The difference of two squares

Worked example 43. Question: Factorise x²−64. Method: Both terms are squares. Using A²−B²=(A−B)(A+B), the expression becomes (x−8)(x+8). Check: Multiplication gives x²+8x−8x−64 = x²−64. Watch for: An expression x²−64 is not the same as (x−8)².

Grouping four terms

Worked example 44. Question: Factorise 6x+7x+36y+42y. Method: Group the x and y terms: (6+7)x+(36+42)y = 13x+78y. Extract the common factor 13: 13(x+6y). Check: Expanding the factorised form gives the same combined coefficients 13 for x and 78 for y. Watch for: Grouping works because the same complete factor is shared; it is not arbitrary rearrangement.

Try it differently: Regroup as 6(x+6y)+7(x+6y) and verify the same final product. Hide the method and reproduce one legal line independently. Being able to start unaided matters more than recognising the printed answer. Use a numerical counterexample to explain why one tempting identity is invalid.

A repeated bracket from grouping

Worked example 45. Question: Factorise 6x+6y+8x+8y. Method: The terms can be grouped as 6(x+y)+8(x+y). Extract the repeated (x+y) bracket to obtain (6+8)(x+y)=14(x+y). Check: Expanding gives 14x+14y, equal to the original collected like terms. Watch for: Do not cancel or drop one repeated bracket; it appears as a common factor.

A greatest obvious common factor

Worked example 46. Question: Factorise 7x²+42x. Method: Each term contains 7x. Factor it out to leave 7x(x+6). This is an algebraic product, not a sum of isolated cancelled terms. Check: Expanding 7x(x+6) gives 7x²+42x, the original expression. Watch for: Taking out x without the numerical coefficient is correct but may leave a form that is not completely factored.

Try it differently: Use the two factors to solve 7x²+42x=0. Ask whether a common factor could be removed before applying the more complicated identity. Explain why the same expression may have several equivalent-looking but differently ordered factor products.

Expand two different simple brackets

Worked example 47. Question: Expand (x+6)(x+9). Method: The four products are x², 6x, 9x and 54; combining like terms gives x²+15x+54. Check: At x=0 both forms equal 54, so at least the constant term is checked independently. Watch for: The coefficient of x is the sum 15, not the product 54.

Try it differently: Reverse this result: factorise the expanded polynomial without looking at the original brackets. Write the first likely sign error in a small correction log and name the operation that prevents it. Substitute x=0 and another value into both forms to catch a coefficient or constant mistake.

Recognise a perfect square in reverse

Worked example 48. Question: Factorise x²+12x+36. Method: The first and last terms are x² and 6². The middle term is twice the product x×6, so the expression is (x+6)². Check: Expanding the answer restores all three original terms in the same order. Watch for: Two squared end terms alone are not enough; verify the middle coefficient exactly.

Try it differently: Write a corresponding expression for (x−6)² and explain its sign change. Compare this pattern with the previous one: which term or coefficient tells you the method should change? Ask whether any remaining factor can be extracted to make the answer more completely factorised.

Non-unit squares in a difference

Worked example 49. Question: Factorise 36x²−81. Method: The two squares are (6x)² and 9². Their difference factors as (6x−9)(6x+9). Check: On expansion, the opposite cross-products −54x and +54x cancel. Watch for: The factors involve 6x, because the coefficient of x² is 36, not 1.

Try it differently: Solve 36x²−81=0 and check both roots. Return after two days without a worksheet heading; decide whether to expand, factorise or group based on the expression. Construct a new example with different coefficients and test it at x=1.

Full distribution with non-unit coefficients

Worked example 50. Question: Expand (6x+6)(7x+8). Method: Multiply all four pairs: 42x² + 48x + 42x + 48. Combine the middle terms to obtain 42x²+90x+48. Check: At x=0 the product form and expanded form both equal 48, and the leading x² coefficient is 6×7=42. Watch for: Failing to include both cross-products changes the middle coefficient.

Try it differently: Check the result by expanding with the two brackets written in reversed order. Give the solution a final reverse check using expansion, not only the fact that it looks neat. Use a numerical counterexample to explain why one tempting identity is invalid.

Five mistakes that a good correction should prevent

Missing the double cross-product

Write two complete brackets and multiply each pair. The two mixed products are why a squared sum includes 2ab. Find the first mistaken operation on the learner’s original paper, then test the correction on a new problem where the coefficients differ.

Assuming a negative bracket creates a negative square

Keep the minus sign during multiplication. A negative term squared creates a positive last term. Find the first mistaken operation on the learner’s original paper, then test the correction on a new problem where the coefficients differ.

Confusing a squared difference with two-square subtraction

(a−b)² is a²−2ab+b², while a²−b² is (a−b)(a+b). These are different relationships. Find the first mistaken operation on the learner’s original paper, then test the correction on a new problem where the coefficients differ.

Grouping terms that share no complete bracket

Group only when lawful factoring yields the same whole factor. Expand the proposed product to check. Find the first mistaken operation on the learner’s original paper, then test the correction on a new problem where the coefficients differ.

Leaving a factorisation unchecked

A valid answer returns to the original polynomial, coefficient by coefficient, when multiplied. Find the first mistaken operation on the learner’s original paper, then test the correction on a new problem where the coefficients differ.

How tutoring can make algebra less fragile

A Secondary 3 G3 Mathematics tutor in Bukit Timah should help students recognise structure, not merely reproduce an expansion demonstration. An effective starting diagnostic could show (x+3)², x²−9 and x²+6x+9. The student explains why the three expressions call for different operations despite their visual similarities.

A small-group setting can support this learning when each pupil completes independent written working and can explain a choice. The tutor checks the earliest incorrect line, models a correction and asks the student to verify by reversing the operation. A later question without hints reveals whether the pupil has actually learned the distinction.

The unchanged Clementi Secondary 1 Mathematics small-group tutorial article illustrates an earlier foundation stage. For the year-level context, see Secondary 3 G3 Mathematics tuition in Bukit Timah. Teaching arrangements and syllabus sequencing should always be confirmed for the individual learner.

A six-week plan for stronger expansion and factorisation

Week 1 — diagnose which direction is missing

Use a small mixed set to see whether the learner can expand but not factorise, or factorise but not explain the identity. Keep a dated first attempt and a later independent attempt. The improvement to watch is method selection and correct reverse checking, not only the number of problems finished.

Week 2 — make bracket multiplication secure

Rebuild distribution with two brackets and write each of the four products, including signs, before collecting terms. Keep a dated first attempt and a later independent attempt. The improvement to watch is method selection and correct reverse checking, not only the number of problems finished.

Week 3 — contrast three identities

Place a sum square, difference square and difference of two squares together. Explain which features distinguish them. Keep a dated first attempt and a later independent attempt. The improvement to watch is method selection and correct reverse checking, not only the number of problems finished.

Week 4 — extract common factors and group

Practise the simplest common factor first, then move to four-term expressions where a whole bracket repeats. Keep a dated first attempt and a later independent attempt. The improvement to watch is method selection and correct reverse checking, not only the number of problems finished.

Week 5 — mix with quadratic equations

Use factorisation as a way to solve zero-product equations and connect with x-intercepts where appropriate. Keep a dated first attempt and a later independent attempt. The improvement to watch is method selection and correct reverse checking, not only the number of problems finished.

Week 6 — retest unseen structure

Change the coefficient sizes and order of terms. Compare the student’s explanations and independent checking with Week 1. Keep a dated first attempt and a later independent attempt. The improvement to watch is method selection and correct reverse checking, not only the number of problems finished.

Frequently asked questions

Why is (a+b)² not just a²+b²?

The complete product has two cross-products, ab and ba, giving the additional 2ab. Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

Why does (a−b)² end with positive b²?

Multiplying two negative b terms produces a positive b². Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

What is the difference of two squares rule?

a²−b²=(a−b)(a+b), with the cross-products cancelling on expansion. Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

What makes grouping possible?

Each useful group must reveal one identical complete factor, often a bracket. Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

How do I check a factorisation?

Expand all proposed factors and compare every original coefficient. Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

Which step comes first in a long algebraic expression?

Check for a common factor before attempting more elaborate identities or quadratic factorisations. Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

Is a factorisation an identity or a solution?

The two sides are equivalent expressions; setting their product equal to zero becomes a separate equation-solving task. Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

Can parents help without solving it all?

Ask which pattern the learner recognises and request an expansion check of the final factors. Try one varied question afterwards to confirm that the explanation is available without copying a worked example.

The 2027 SEAB G3 syllabus index and the K310 Mathematics syllabus remain authoritative for the examination framework. This guide is a learning companion, not a replacement for school-specific instructions.

The takeaway

Did you know? Being able to expand a product and then recover it by factorisation is one of the simplest ways to show genuine algebraic understanding. Identify the structure, work carefully with brackets, and verify every coefficient. A tutor’s best contribution is helping the learner make those decisions independently when the next question looks unfamiliar.