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The difference of two squares, a² − b², means square each quantity and then subtract. The square of their difference, (a − b)², means subtract first and then square the result. Those instructions generally produce different values, so the brackets cannot be ignored.
For G2 Mathematics, keep the two identities separate: a² − b² = (a − b)(a + b), while (a − b)² = a² − 2ab + b². The second expression has a middle term because both cross-products appear when the same bracket is multiplied by itself.
If your child writes a² − b² = (a − b)², the repair is to expand the brackets and compare the operations. The learner may recognise the words square and difference without yet controlling their order. A small structural distinction makes many later factorisation questions clearer.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–29
With a = 5 and b = 3, the difference of squares is twenty-five minus nine, or sixteen. The square of the difference is (5 − 3)² = 2² = 4. The two results are not equal.
The correct factorisation of twenty-five minus nine is (5 − 3)(5 + 3) = 2 × 8 = 16. One factor is the difference and the other is the sum. Replacing the sum factor with another difference changes the value.
This numerical example disproves the proposed universal equality, but the general explanation comes from expansion. The cross terms cancel in (a − b)(a + b); they combine in (a − b)(a − b). That is the mechanism behind the two different identities.
This guide uses real-number variables unless otherwise stated. Geometric area examples use nonnegative lengths with the larger square identified appropriately. Algebraic identities themselves do not require the variables to be positive or one to be larger than the other.
The SEAB G2 Mathematics syllabus for 2027 includes expansion, factorisation and the standard square and difference-of-squares relationships. Understanding their structure helps students choose the correct form rather than matching a few familiar symbols by appearance.
These relationships support quadratic equations, algebraic fractions and geometric calculations. A mistaken identity can spread into several later steps, even when those steps are performed consistently. It is therefore useful to repair the original expression before reviewing every downstream answer.
The broader factorisation guide explains common factors and quadratic structure. This page focuses on the particular confusion between subtracting squares and squaring a subtraction.
G2 describes a subject level, not a learner’s intelligence or future. A student who has confused these two patterns can improve through one clear expansion and several contrasting examples. Start with the operation order and the meaning of the brackets, then build speed after the relationship is secure.
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In a² − b², the powers apply separately to a and b. First calculate a multiplied by itself and b multiplied by itself, then subtract the second result from the first. There are no brackets placing the entire difference under one square.
In (a − b)², the exponent applies to the complete bracket. Calculate the difference and multiply that difference by itself. The notation means (a − b)(a − b), not a² − b².
Reading aloud can reveal the distinction: “the difference of the squares” versus “the square of the difference.” The order of the words reflects the order of the operations. Neither phrase is a stylistic alternative for the other.
The algebra-instruction guide develops this connection between language and mathematical action. Before factorising or simplifying, identify which complete quantity the power affects. A correctly read expression is the foundation for every valid transformation that follows.
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Multiply (a − b)(a + b) term by term. The four products are a², ab, −ab and −b². The middle two terms cancel because ab plus negative ab is zero.
The remaining expression is a² − b². This establishes the identity for every real a and b. No particular numerical values were chosen, and no division introduced a nonzero restriction.
The cancellation is the feature to notice. The two brackets share the same terms but have opposite signs between them. One cross-product is positive and the other negative, so they disappear when combined.
Students can verify a proposed factorisation by expanding it back. If the original expression does not return, the factorisation is not equivalent. This check is more reliable than asking whether the factors resemble a memorised pattern, because it tests the actual value-preserving relationship rather than a visual impression.
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Now multiply (a − b)(a − b). The four products are a², −ab, −ab and b². The two middle terms have the same negative sign, so they combine to −2ab rather than cancel.
The final term is positive because negative b multiplied by negative b gives b². Therefore (a − b)² = a² − 2ab + b². The signs of both the middle term and the final term matter.
Comparing this with the previous section explains the entire distinction. Changing the second bracket from a + b to a − b changes two of the products. It is not a harmless alteration of notation.
A learner who writes a² − b² after expanding should identify all four products separately before combining like terms. That temporary scaffold makes the missing cross-products visible. Once the structure is understood, the standard identity can compress the working without concealing the reason it is true.
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A compact table helps keep the expanded form and factor structure attached to the correct expression. Use it as a meaning check rather than a list to recite without expansion.
| Starting expression | Equivalent form | What happens to cross-products? |
|---|---|---|
| a² − b² | (a − b)(a + b) | They cancel when the product is expanded |
| (a − b)² | a² − 2ab + b² | They combine negatively |
| (a + b)² | a² + 2ab + b² | They combine positively |
The final two rows each describe one bracket repeated. The first describes two related but different brackets. That structural distinction is more dependable than trying to infer the answer from the presence of two square symbols alone.
When checking a student’s result, ask which row matches the original expression before reviewing the arithmetic. A correct identity used on the wrong structure still changes the problem. The goal is to recognise why a pattern applies, not simply remember that all three formulas contain a² and b².
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For real a and b, (a − b)² is always nonnegative. By contrast, a² − b² can be negative. With a = 3 and b = 5, it is nine minus twenty-five, or negative sixteen.
The square of the difference is (3 − 5)² = (−2)² = 4. This sign contrast is a quick way to reject the proposed identity in that example. A negative value cannot equal a real square.
Do not overgeneralise the check. A nonnegative difference of squares is not automatically equal to the square of the difference. The opening example produced sixteen and four, both nonnegative but unequal.
The sign test is therefore a useful error detector with a limited role. Failure decisively rejects an answer; passing it does not establish equivalence. Use the general expansion to justify an identity and numerical or sign checks to expose particular mistakes efficiently.
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Take a = −5 and b = 3. The difference of squares remains twenty-five minus nine, or sixteen. The correct factors are (−5 − 3)(−5 + 3) = (−8)(−2) = 16.
The square of the difference is (−5 − 3)² = 64. The two expressions still differ. The factorisation identity works because its expansion is valid for all real inputs, not because a and b are lengths or positive numbers.
This is a useful distinction between algebra and a geometric illustration. A diagram showing positive square side lengths can explain one case, but the algebraic proof covers negative values as well. Do not import a length restriction into an unrestricted algebra question.
When substituting negative numbers, use brackets around the substituted value. Writing (−5)² makes it clear that the complete negative number is squared. A notation error involving the placement of a minus sign can otherwise obscure the identity issue being tested.
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Although a² − b² and (a − b)² are not identical, they can agree for particular inputs. If b = 0, both equal a². If a = b, both equal zero. These successful cases do not establish a universal identity.
To find all real cases of equality, compare the expansions. Setting a² − b² equal to a² − 2ab + b² gives 2b(a − b) = 0. Thus b = 0 or a = b.
This explains why checking only zero or equal inputs can hide the mistake. Those choices remove precisely the difference between the expressions. A useful counterexample should choose values outside those special cases.
The identity-versus-equation guide develops this scope distinction. An identity must hold throughout its stated domain, while an equation can select particular values. A few successful substitutions should be described as case checks, not promoted into proof of a general relationship.
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For positive lengths a = 5 and b = 3, a large square has area twenty-five and a smaller square area nine. Removing the smaller square’s area leaves sixteen square units. This represents a² − b².
A square whose side is the difference, two, has area four. It is a different region. Subtracting a side length before forming a square does not remove the same area as subtracting a complete smaller square afterward.
The remaining L-shaped region from a corner removal can be rearranged into a rectangle with side lengths a − b and a + b when the standard construction is made. For five and three, that rectangle is two by eight, giving sixteen.
The model is an illustration under the condition a ≥ b ≥ 0, not a restriction on the algebraic identity. If the variables can be negative, use expansion for the general proof. Keep the geometric quantities and the unrestricted algebraic symbols connected without pretending that every real input is a physical side length.
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To calculate 101² − 99², factor it as (101 − 99)(101 + 99). The factors are two and two hundred, so the result is four hundred. This avoids calculating two large squares separately.
The square of their difference would be (101 − 99)² = 4, which is dramatically different. The fact that the inputs are close does not justify discarding the sum factor. In this example, the sum is the large factor that makes the result four hundred.
Another example is 52² − 48² = 4 × 100 = 400. The common difference of four and sum of one hundred expose the structure. A mental shortcut is valid because it preserves the identity exactly.
Students should still read the requested expression carefully. If the question actually asks for (101 − 99)², the answer is four. A useful method answers the given operation order rather than replacing it with whichever pattern seems easiest to calculate.
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The expression 4x² − 25 is a difference of squares because 4x² = (2x)² and 25 = 5². Its factorisation is (2x − 5)(2x + 5). The square roots of the complete terms form the two bracket entries.
Writing (4x − 25)(4x + 25) would square the wrong quantities on expansion, producing 16x² − 625. The coefficients and constants cannot be copied directly without considering which quantities were squared.
Similarly, 9a² − 16b² factors as (3a − 4b)(3a + 4b). Identify each complete square first, then apply the difference-and-sum pattern. This step makes the method reliable when letters and numerical coefficients appear together.
Check by expanding one example. The cross-products −12ab and +12ab cancel, leaving 9a² − 16b². The check confirms both the square roots chosen for the factors and the opposite signs needed for cancellation.
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The expression 2x² − 18 has a common factor of two. Writing it as 2(x² − 9) reveals a difference of squares inside the bracket. The complete factorisation is 2(x − 3)(x + 3).
The factor two remains outside. It does not disappear when the inner expression is factorised. Expanding the final product gives 2x² − 18, confirming that the original value is preserved.
This example connects two decisions: look for a common factor, then inspect the remaining structure. A learner who searches only for a directly visible pair of squares may miss an easy first step. A learner who extracts the common factor but stops may leave a further requested factorisation incomplete.
The wording of the task matters. If it asks to factorise completely over the usual real or integer-coefficient school setting, continue as appropriate. If it asks only for a particular equivalent form, follow that instruction. Algebraic fluency includes recognising a structure and knowing which final form the question requests.
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The expression 9 − x² factors as (3 − x)(3 + x). It is not the same as x² − 9, which factors as (x − 3)(x + 3). The subtraction order matters.
One can also write 9 − x² = −(x² − 9) = −(x − 3)(x + 3). The negative sign records the reversed difference. Since 3 − x = −(x − 3), the two correct forms agree.
At x = 5, 9 − x² is negative sixteen. The factors (3 − 5)(3 + 5) give negative two times eight, also negative sixteen. This check exposes a lost sign immediately.
Do not reorder a subtraction merely to make the first term look more familiar. Addition permits reordering without a sign change; subtraction does not. A correct identity applied after an incorrect reorder will faithfully factorise the wrong expression, so preserve the original order before choosing the pattern.
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The expression x² + 9 is not a difference of squares. Multiplying (x − 3)(x + 3) gives x² − 9, so those factors do not represent x² + 9 over the real numbers.
For real x, x² + 9 is strictly positive. It therefore cannot have a real zero, whereas (x − 3)(x + 3) is zero at three and negative three. Their different zero behaviour is another way to see that the expressions are not equivalent.
This guide does not need complex-number factorisation. In the ordinary real-number setting, keep the distinction clear and do not apply the difference-of-squares pattern merely because two squared-looking terms are present.
Likewise, (a + b)² is not a² + b² in general. Its expansion contains the middle term 2ab. The presence of a plus sign changes which standard relationship applies, but it does not remove cross-products from a repeated bracket. Read the complete structure before using any shortcut.
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Compare x² − 9 = 0 with (x − 3)² = 0. The first factors as (x − 3)(x + 3) = 0, giving x = 3 or x = −3. The second has a repeated factor x − 3 and gives only x = 3.
The different solution sets are a consequence of the different expressions. A false replacement of x² − 9 by (x − 3)² would lose the valid solution negative three from the first equation.
Check the omitted value in the originals. At x = −3, x² − 9 is zero, while (x − 3)² is thirty-six. This is a decisive check of the alleged equivalence.
Remember that factorising an expression is not automatically solving an equation. The zero condition in these examples is explicitly supplied. If the task asks only to factorise x² − 9, report its factors without inventing an equation or treating the two root values as the factorised expression.
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A. Evaluate 7² − 4² and (7 − 4)² separately. B. Expand (x − 5)(x + 5). C. Expand (x − 5)². D. Factorise 9x² − 4.
E. Factorise 3x² − 27 completely. F. Factorise 16 − y². G. Calculate 202² − 198² efficiently. H. Solve x² − 16 = 0 and compare its solutions with (x − 4)² = 0.
I. Simplify (x² − 9)/(x − 3), stating the original restriction. J. Explain why x² + 16 does not factor as (x − 4)(x + 4) over the real numbers. K. Give a counterexample to a² − b² = (a − b)².
For L, find all real pairs a and b for which that equality does hold. For M, simplify (x + 2)² − (x − 1)² without fully expanding both squares first. State which complete expressions take the roles of the two terms being squared.
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A gives forty-nine minus sixteen, or thirty-three, while the squared difference is nine. B gives x² − 25. C gives x² − 10x + 25. D factors as (3x − 2)(3x + 2).
E is 3(x − 3)(x + 3). F is (4 − y)(4 + y). G is (202 − 198)(202 + 198) = 4 × 400 = 1,600. H has solutions four and negative four for the difference-of-squares equation, but only four for the repeated-square equation.
I simplifies to x + 3 with x ≠ 3. J fails because those proposed factors expand to x² − 16. K can use a = 5, b = 3, producing sixteen and four. For L, equality holds when b = 0 or a = b.
For M, use complete terms x + 2 and x − 1. Their difference is three and their sum is 2x + 1, so the result is 3(2x + 1) = 6x + 3. Replacing the original expression by the square of three would incorrectly give nine for every x.
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The fraction (x² − 9)/(x − 3) is undefined at x = 3. Factorising the numerator gives (x − 3)(x + 3), allowing cancellation only when x − 3 is nonzero. The simplified expression is x + 3 for x ≠ 3.
By contrast, (x − 3)²/(x − 3) simplifies to x − 3 on the same restricted domain. The two numerators do not become interchangeable merely because both contain a factor x − 3 after correct factorisation.
The fractional-equation validity guide develops this domain safeguard. Simplification preserves values where the original expression is defined; it does not restore an excluded denominator value.
When reviewing a solution, check the identity used before checking cancellation. A student may cancel correctly from a wrongly factorised numerator. Repairing only the cancellation rule would miss the earlier structural error that changed the entire expression.
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The expression √(a² − b²) is not generally a − b. With a = 13 and b = 5, the radicand is 169 − 25 = 144, so the square root is twelve. The difference thirteen minus five is eight.
This example often appears naturally in Pythagoras’ theorem: a right triangle with hypotenuse thirteen and one leg five has another leg twelve. Squaring the side difference would solve a different problem.
The radicand must be nonnegative for a real square root. When side lengths are involved, the hypotenuse is at least as long as either leg, so the relevant subtraction order matters. Do not reverse it or assume every symbolic input creates a valid real length.
Even √((a − b)²) equals the absolute value of a − b over the real numbers, not always a − b itself. A square root returns the nonnegative root. This further distinction reinforces why powers, roots and brackets need to be read as complete operations rather than distributed casually across every sign.
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In (x + 2)² − (x − 1)², the two squared quantities are the complete brackets x + 2 and x − 1. Apply the identity to those complete expressions before simplifying their sum and difference.
The difference bracket is (x + 2) − (x − 1) = 3. The sum bracket is (x + 2) + (x − 1) = 2x + 1. Their product is 6x + 3.
Writing temporary names A = x + 2 and B = x − 1 can make the pattern clearer: A² − B² = (A − B)(A + B). Substitute back carefully, preserving brackets until the subtraction has been distributed correctly.
This is an optional extension for a learner secure with the basic identity. It shows that the pattern concerns mathematical structure rather than the particular letters a and b. The same relationship applies when each squared term is itself a more complicated expression, provided every transformation remains valid on the stated domain.
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One direct way to test an alleged identity is to subtract one side from the other and simplify. Here a² − b² minus (a − b)² becomes 2ab − 2b², or 2b(a − b).
This difference is not always zero, so the expressions are not identical. Its zero cases give the special equality conditions already found: b = 0 or a = b. The calculation connects a counterexample with a general structural explanation.
Be careful with the outer subtraction. The entire expanded square must be subtracted, including its positive b² term. Losing those brackets can introduce a new sign error while trying to diagnose the original one.
The method is useful beyond this pair, but it should not become unnecessary work for every simple expansion. Use it when a general equality is in doubt or when the question asks when two expressions coincide. Choose a check that resolves the actual uncertainty rather than repeating algebra without a clear purpose.
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SECTION 23 OF 29
22. Preserve coefficient and sign information in a calculator check
To compare the expressions numerically, enter complete brackets around substituted negative values and around the difference being squared. The inputs 5² − 3² and (5 − 3)² should represent the two distinct operations, not two accidentally identical key sequences.
A calculator can verify a chosen counterexample but cannot turn a finite set of successful checks into a general proof. Inputs such as b = 0 can conceal the missing term, so choose a case that distinguishes the structures.
Keep the exact small-number results when possible. Sixteen and four make the difference obvious without rounding. Approximate displays can sometimes conceal small algebraic discrepancies in other examples, so matching rounded values should not be treated as proof of equivalence.
The guide to estimate checks supports this distinction between rejecting a claim and establishing it. A counterexample can disprove a universal identity; a general value-preserving derivation establishes the correct one.
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If a student replaces a² − b² with (a − b)², repair the identity choice. If they choose (a − b)(a + b) correctly but expand a cross-product with the wrong sign, repair that multiplication. If they factorise accurately but solve an expression with no equation, repair the task interpretation.
If they omit x ≠ 3 from a fraction, repair the domain. If they lose negative three from x² − 9 = 0, inspect whether a repeated square was substituted incorrectly or a factor was divided away.
These errors call for different practice. Ask the learner to explain the original expression in words and then show the four products in one expansion. That usually reveals whether the difficulty lies in reading, structure, signs or final interpretation.
Keep feedback specific. “The cross terms cancel only when the brackets have opposite signs” gives a concrete decision to remember. “Learn your identities” may be true as a broad goal, but it does not explain the mechanism needed to repair the next attempt independently.
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Begin with a² − b² and (a − b)² using the same small numerical values. Then compare their expansions. Next vary one sign, one coefficient or one complete bracket at a time so the student can see which feature changes the structure.
A useful set is x² − 9, (x − 3)² and x² + 9. Ask for the correct interpretation of each before factorising or expanding. This prevents a learner from applying the difference-of-squares pattern merely because x² and nine are visible.
The worked-example revision guide supports predicting a step before reading it and gradually removing assistance. Here the prediction should name the structure, not just guess a final bracket form.
After one successful supported example, use changed values independently. The learner should be able to explain why one pair of cross-products cancels and another combines. That explanation is stronger evidence of understanding than reciting a formula while remaining unsure which original expression it applies to.
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This distinction supports quadratic factorisation, equation solutions, algebraic fractions and right-triangle calculations. The same bracket-reading habit also helps when rearranging formulas or simplifying expressions with complete sums inside powers.
Do not rush to every application at once. If the learner is still uncertain about expansion, keep the examples small. Once the four products are secure, introduce a coefficient such as 4x² − 25, then an equation or fraction with its additional conditions.
The Clementi Secondary 1 Mathematics small-group guide gives the wider teaching context. Effective support identifies the earliest missing relationship and checks it independently before adding more layers of notation.
The goal is a student who recognises what is squared, preserves the operation order and verifies a new factorisation by expansion. Speed can follow from that control. Memorising two similar-looking answers without their different mechanisms leaves the original confusion ready to return in a less familiar question.
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For a nonnegative integer n, compare the consecutive squares (n + 1)² and n². Their difference factors as ((n + 1) − n)((n + 1) + n), giving 1 × (2n + 1) = 2n + 1.
The difference between the original numbers is always one, but the difference between their squares grows with n. For ten and eleven, the squares are one hundred and one hundred twenty-one, with a difference of twenty-one. The square of the number difference is only one.
This makes the false identity particularly visible. Replacing a difference of consecutive squares with a squared difference would erase the dependence on n and predict the same answer for every pair. The correct sum factor retains the information about how large the original numbers are.
The familiar sequence of positive odd gaps between consecutive nonnegative squares follows directly: one, three, five, seven and so on. Each gap is 2n + 1 for its starting n. The pattern is not evidence that squaring distributes over subtraction; it is evidence of the correct difference-and-sum factorisation.
For a fixed positive step k, the same reasoning gives (n + k)² − n² = k(2n + k). A fixed difference k between the inputs still produces different squared differences as n changes. Both the gap and the sum matter.
For example, numbers eight and ten differ by two, and their squares differ by thirty-six. Numbers eighteen and twenty also differ by two, while their squares differ by seventy-six. The squared input gap is four in both cases, so it cannot describe either original difference of squares.
This optional extension helps a learner move beyond isolated counterexamples. It explains a whole family of cases using one general expression. The student can predict the change before calculating individual squares, then verify a small numerical pair independently.
Keep the stated domain clear when discussing an increasing sequence of square values. The nonnegative-integer assumption makes that ordering immediate. The underlying algebraic identity is valid more broadly, but conclusions about which square is larger still depend on the inputs being compared.
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SECTION 28 OF 29
27. A known difference of squares does not fix the input difference
Suppose positive numbers a and b satisfy a² − b² = 36, with a greater than b. It is tempting to take a square root and conclude that a − b = 6. That conclusion would replace the given expression with a different one.
The pair a = 10 and b = 8 satisfies the original condition because one hundred minus sixty-four is thirty-six. Its input difference is two, not six. Another positive pair, a = 6.5 and b = 2.5, has squares 42.25 and 6.25, again differing by thirty-six, while its input difference is four.
Both pairs meet the same difference-of-squares condition but have different differences between the inputs. Therefore the original information does not determine one unique a − b. This is a structural information issue, not a calculator-accuracy issue.
The correct identity says (a − b)(a + b) = 36. Knowing the product of these two factors does not identify either factor without additional information. The factor representing the sum is precisely what the false square replacement loses.
If the question also gives a + b = 18, then a − b = 36/18 = 2. The two linear relationships yield a = 10 and b = 8. The extra sum information has now fixed the difference and the individual values.
If the sum is nine instead, the difference is four, giving a = 6.5 and b = 2.5. The two examples reconcile through the same product relationship. They do not represent two competing answers to a fully specified problem; they represent different possibilities under incomplete initial information.
A clear student explanation is, “The square difference fixes the product of the sum and difference, so the difference alone is not yet known.” That sentence connects the identity to the reason more information is needed. It is stronger than merely saying that the proposed answer six is wrong.
The insufficient-information guide develops this two-valid-examples test. Here it reinforces the importance of preserving the sum factor when reasoning backward from a difference of squares.
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Can the two expressions ever have equal values?
Yes. Over the real numbers they agree when b = 0 or a = b. Those special cases do not make the expressions identical. A universal identity must hold for every allowed pair, not just selected pairs that remove the difference.
Why does the difference-of-squares product use opposite signs?
Opposite signs make the cross-products cancel. Repeating the same difference bracket makes two negative cross-products combine instead. Expanding all four products shows the mechanism directly.
Is a² + b² the square of a + b?
Not generally. The square of a + b includes the additional term 2ab. The same care with complete brackets applies to both addition and subtraction.
Can I cancel a square term from a subtraction?
Cancellation applies to common factors, not arbitrary terms inside a sum or difference. For a ≠ 0, the fraction (a² − b²)/a² equals 1 − b²/a². It does not become 1 − b² merely because a² appears in the numerator and denominator. With a = 5 and b = 3, the correct value is sixteen twenty-fifths. Writing the numerator as (a − b)(a + b) can reveal genuine factors, but those factors must actually match a denominator factor before cancellation is allowed. Preserve the complete expression and its original nonzero restrictions throughout.
Which idea should come first?
Read the operation order and identify the complete quantities under the powers. Then use expansion to justify the chosen identity. The Mathematics explanation directory connects this guide to related work on algebra instructions, domains and checking.