Recognising an expression as an identity in disguise

What to assume they know

• The identities (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b², and how each was obtained (An identity holds for every value; an equation need not, Reading (a + b)² off a partitioned square)
• Taking a numerical common factor out of a sum of terms
• Square roots of perfect squares, and recognising that 50 has no whole-number square root (Proof by contradiction: why √2 cannot be a ratio of integers)
• Squaring a product such as 6x or 5p, and knowing (6x)² = 36x²
• Multiplying and dividing by the same non-zero number leaves an expression unchanged

What they should be able to do

• Given a three-term expression, propose values for a and b by inspecting the first and last terms
• Test the proposal against the middle term and decide whether the expression is the square of a binomial
• Factorise expressions whose leading coefficient is a perfect square, such as 36x² + 12x + 1
• Recognise when the leading term's square root is irrational, and take out a common factor so that the identity can still be applied
• Apply the same move when the common factor to be removed is a fraction rather than a whole number
• Use the minus form of the identity on expressions whose middle term is negative, including ones written with the terms out of order
• Produce an expression that fails the middle-term test, and state what that failure rules out
• Distinguish the statement "(x + 2) is a factor" from the stronger statement that the expression equals (x + 2)²
• State the class of expressions this method reaches, and name what is needed for the rest

Where it usually goes wrong

• "Factorising is guessing until something works." Not here. The first and last terms leave no room for choice; only the middle term can decide, and it decides yes or no. Turn the whole procedure into a two-step check.
• "36x² means a = 36x." It means a² = 36x², so a = 6x. Students carry the coefficient into a instead of taking its square root. Example 6 exists for exactly this.
• "If √50 shows up, the expression cannot be factorised." It can — the radical is a signal to look for a common factor, not a dead end. The chapter prints √50 p and then routes around it.
• "You can only take out a whole number as a common factor." The two starred exercises need 1/3 and 1/5. Removing a fraction is the same operation.
• "Every three-term quadratic is a perfect square." Most are not. Run the middle-term test on x² + 5x + 4 and let it fail in front of the student.
• "A negative middle term means the last term is negative too." In (a − b)² the last term is +b². 16y² − 24y + 9 has a positive 9.
• "The terms must be written in the order a², 2ab, b²." 16s² + 25t² − 40st is the same expression with the middle term last. Reordering is the first move, not a different problem.
• "(x + 2) is a factor" and "the expression is (x + 2)²" say the same thing. The second implies the first and not the other way round.

Questions to check understanding

• Factorise a three-term expression by identifying a and b, showing the middle-term check
• Factorise completely, where a numerical or fractional common factor must come out before the identity applies
• Given a three-term expression, state whether it is the square of a binomial and justify the answer either way
• Fill a missing coefficient so that a given expression becomes a perfect square
• Factorise an expression whose terms are printed out of the usual order
• Name one factor of an expression, and separately write the expression in fully factorised form

Examples worth working on the board

Inputs, not answers. Values marked Verified are worked out here; this book prints no answer key.

• Example 5 (§4.3, p. 72). Factorise x² + 4x + 4. The page walks the matching explicitly: x² is the square of x, 4 is the square of 2, and the middle term 4x is twice the product of x and 2. So a = x and b = 2. Verified: the expression is (x + 2)². Note: the page states the three observations before it names a and b, which is the correct order — the naming is a conclusion, not a guess.
• Example 6 (§4.3, p. 72). Factorise 36x² + 12x + 1, with a = 6x and b = 1. Verified: 2(6x)(1) = 12x, so the test passes and the expression is (6x + 1)². This is the first case where a is not just the variable, and it is worth pausing on 36x² = (6x)², since students reliably read the leading coefficient as belonging to a² rather than to a.
• Example 7 (§4.3, pp. 72–73). Factorise 50p² + 60pq + 18q². The page asks what a and b are here, observes that the quantity whose square is 50p² is √50 p — the radical is printed on p. 72; it simply does not survive text extraction — and then says it would rather avoid the radical sign. So it takes 2 out as a common factor: 50p² + 60pq + 18q² = 2(25p² + 30pq + 9q²), and inside the bracket a = 5p and b = 3q. Verified: 2(5p)(3q) = 30pq, so the test passes and the answer is 2(5p + 3q)².
• Why taking out 2 works at all (§4.3, p. 73). The reason to narrate: 50 is not a square but 25 is, and 50 = 2 × 25. Removing the non-square part of the leading coefficient is what leaves a square behind. The chapter states the move and not the reason, so the reason is added here.
• Exercise Set 4.2 Q1 (pp. 74–75), six items. Unstarred: 9x² + 24xy + 16y²; 4s² + 20st + 25t²; 49x² + 28xy + 4y²; 64p² + (32/3)pq + (4/9)q². Starred, with a printed hint pointing back at Example 7's common-factor move: 3a² + 4ab + (4/3)b² and (9/5)s² + 6sv + 5v². Verified: 9x² + 24xy + 16y² = (3x + 4y)² since 2 × 3 × 4 = 24; 4s² + 20st + 25t² = (2s + 5t)² since 2 × 2 × 5 = 20; 49x² + 28xy + 4y² = (7x + 2y)² since 2 × 7 × 2 = 28; 64p² + (32/3)pq + (4/9)q² = (8p + (2/3)q)² since 2 × 8 × (2/3) = 32/3; 3a² + 4ab + (4/3)b² = (1/3)(9a² + 12ab + 4b²) = (1/3)(3a + 2b)²; (9/5)s² + 6sv + 5v² = (1/5)(9s² + 30sv + 25v²) = (1/5)(3s + 5v)². The pedagogically important difference: in Example 7 the factor removed is the whole number 2, whereas in the two starred items it is 1/3 and 1/5. The hint invites the same move with a fractional factor, which is a genuine step up.
• The minus form used the same way (two exercise sets, both printed later than this section; note the numbering does not track the sections — Exercise Set 4.3 is §4.4's, on pp. 76–77, while Exercise Set 4.4 comes after §4.6, on pp. 81–82). Exercise Set 4.3 Q2 (p. 77) supplies 16y² − 24y + 9 and p²/16 − 2 + 16/p²; Exercise Set 4.4 Q3 (p. 82) supplies 16s² + 25t² − 40st and 49g² + 14gh + h². Verified: 16y² − 24y + 9 = (4y − 3)²; p²/16 − 2 + 16/p² = (p/4 − 4/p)², where the middle term is −2 because 2 × (p/4) × (4/p) = 2 exactly; 16s² + 25t² − 40st = (4s − 5t)², which has to be read with its terms reordered before the pattern is visible; 49g² + 14gh + h² = (7g + h)². Note that p²/16 − 2 + 16/p² is the item where students most often stall, because the middle term carries no letter at all.
• A candidate that fails. An added example, not the chapter's: test x² + 5x + 4. The end terms force a = x and b = 2, so the middle term would have to be 4x. It is 5x, so the test fails and x² + 5x + 4 is not the square of a binomial — even though it does factorise, as (x + 1)(x + 4). This is the cleanest way to show that the method is a recogniser and not a general factoriser, and it sets up §4.6.
• The weaker printed conclusion (§4.3, p. 72). Having shown that x² + 4x + 4 = (x + 2)², the page concludes that (x + 2) is a factor of it. Both statements are true and the second is weaker: it drops the information that the factor occurs twice and that there is nothing else. Worth one beat, because examination questions ask for both forms.

Figures to have open

• No figure from the chapter is required. §4.3 carries only Fig. 4.3, which belongs to Reading (a + b)² off a partitioned square; the factorisation work on pp. 72–73 is entirely symbolic.
• A two-slot template — a² in one box, b² in the other, with the middle term shown as a gate the candidate must pass — is worth building. Standard schematic, not in the book.
• A containment diagram for section 11 (squares inside all factorisable quadratics inside all quadratics). Standard schematic.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 4, "Exploring Algebraic Identities": §4.3, whose printed heading begins "Factorisation of Algebraic Expressions Using" and completes as "Identities" on the next line, pp. 72–73, Examples 5, 6 and 7.
• Exercise Set 4.2 Q1 with its printed hint, pp. 74–75.
• Further items of the same type printed later in the chapter: Exercise Set 4.3 Q2 (i), (ii), (iv), p. 77; Exercise Set 4.4 Q3 (ii), (iv), p. 82; End-of-Chapter Exercises Q3 (viii) and Q5 (i), pp. 89.
• Chapter summary, p. 90, third bullet, on using identities to factorise.
• Companion topics: Reading (a + b)² off a partitioned square for where the identity came from, Splitting the middle term once the tiles come away for the quadratics this method cannot reach.