Splitting the middle term once the tiles come away

What to assume they know

• The tile model and what it establishes (Algebra tiles: factorising by rebuilding the rectangle)
• The identity that turns (x + a)(x + b) into x² + (a + b)x + ab
• That two polynomials are equal exactly when their corresponding coefficients are equal (Univariate polynomials and what degree names)
• Signed arithmetic: which pairs of integers multiply to a positive with a negative sum (Why a debt times a debt is a fortune)
• Factor pairs of small whole numbers, and listing them systematically
• That a product of two numbers is zero only when one of them is zero

What they should be able to do

• Compare coefficients of two quadratic expressions and extract the sum and product conditions on a and b
• List the factor pairs of the constant term and select the pair whose sum matches the x-coefficient
• Explain why a pair with the right product but the wrong sum has to be rejected
• Factorise a quadratic whose x-coefficient is negative and whose constant is positive, and say why both parts of the split are negative
• Predict the signs of the two parts from the signs of the constant and the x-coefficient, before doing any arithmetic
• Complete a partially factorised identity, including one where the leading coefficient is not 1
• Split the middle term and factorise by grouping the four resulting terms
• Use an identity to compute a product of two numbers without multiplying them directly
• Turn a worded area problem into a quadratic equation, factorise it, and select the admissible solution
• Justify discarding a solution on the grounds of what the letter stands for

Where it usually goes wrong

• "a + b = 7 and ab = 12 are two rules to memorise." They are one comparison, performed twice: once on the x-coefficient, once on the constant. Show the two expressions stacked and the matching drawn as two arrows.
• "Any factor pair of the constant will do." The chapter itself offers 2 and 15 and then 3 and 10, and rejects both. The product condition narrows the field; the sum condition picks the winner.
• "There are two different factorisations, a = 3, b = 4 and a = 4, b = 3." Same product, brackets written in the other order. Say it once.
• "A negative middle term means one part is negative." Not when the constant is positive: then both are negative. Three sign patterns, and they are decided by the constant first and the middle term second — positive constant with negative middle gives two negatives, negative constant gives one of each.
• "If the tiles cannot show it, the method breaks." Example 12 is precisely the case the printed tile model cannot lay out, and the algebra handles it without comment. That is the argument for leaving the picture behind.
• "With 6x² in front, look at factor pairs of 2." Look at factor pairs of 6 × 2 = 12. The chapter's own Q1 (iii) and (iv) require this and the chapter never says so.
• "x² − 4x − 96 = 0 has two answers, so the pool has two sizes." The equation has two roots; the pool has one length, because a length cannot be −8. Keep the algebraic step and the physical step visibly separate.
• "Rejecting a root is part of solving the equation." It is not — End-of-Chapter Q8 keeps both. What licenses rejection is the meaning of the letter.
• "18 × 29 must be a difference of squares because 23 × 17 was." 18 and 29 have no whole number midway between them. The wider identity covers both, and the difference of squares is its special case.

Questions to check understanding

• Factorise a quadratic with leading coefficient 1, positive or negative constant, showing the factor pairs considered
• Factorise a quadratic with a leading coefficient other than 1, by splitting the middle term and grouping
• Complete a partially given factorisation, supplying the missing bracket or coefficient
• Evaluate a product of two numbers by choosing and naming a suitable identity
• Form a quadratic equation from a worded area or measurement problem, solve it by factorising, and justify which root is kept
• Given that two stated expressions are factors of a quadratic, deduce a relation between its coefficients — the form of End-of-Chapter Q10

Examples worth working on the board

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

• Example 10 (§4.6, p. 81). Set x² + 7x + 12 against x² + (a + b)x + ab and compare the x-coefficient and the constant. Inputs: the two expressions. Verified: a + b = 7 and ab = 12 force a = 3 with b = 4, or the same pair the other way round, and the page says so — the two possibilities are the same factorisation with the brackets swapped, which is worth one sentence so nobody thinks there are two answers.
• Example 11 (§4.6, p. 81). Factorise x² + 11x + 30. The page names two candidate pairs and rejects them: a = 2 with b = 15, and a = 3 with b = 10. Verified: both have ab = 30, but 2 + 15 = 17 and 3 + 10 = 13, neither of which is 11; the pair 5 and 6 satisfies both conditions.
• Example 12 (§4.6, p. 81). Factorise x² − 5x + 6. The page notes first that the x-coefficient is negative. Verified: a + b = −5 with ab = 6 forces a = −2 and b = −3, so the factorisation is (x − 2)(x − 3). The instructive part: the constant is positive, so the two parts share a sign, and the sum is negative, so that shared sign is minus. Neither part is negative "because of the minus in the middle" — both are negative because a positive product and a negative sum leave no other option.
• Exercise Set 4.4 Q1 (p. 81), four fill-in items. (i) s² − 11s + 24 as a product of two brackets. (ii) a missing bracket times (x + 1) giving 3x² − 4x − 7. (iii) 10x² − 11x − 6 written as (2x − __)(__ + 2). (iv) 6x² + 7x + 2 as a product of two brackets. Verified: (i) (s − 3)(s − 8), since −3 − 8 = −11 and (−3)(−8) = 24; (ii) the missing bracket is (3x − 7), since (3x − 7)(x + 1) = 3x² − 4x − 7; (iii) (2x − 3)(5x + 2), so the blanks are 3 and 5x; (iv) (2x + 1)(3x + 2). Items (ii), (iii) and (iv) all have a leading coefficient other than 1, so the plain "factor pairs of the constant" recipe does not apply unchanged — the product to split is the leading coefficient times the constant. For (iii) that is 10 × (−6) = −60, split as −15 and +4; for (iv) it is 6 × 2 = 12, split as 3 and 4. The chapter never states this extension.
• Splitting followed by grouping (§4.8, p. 88, inside Example 18). Once the middle term is split, the four terms are paired and a common bracket is taken out: x² − 12x + 8x − 96 becomes x(x − 12) + 8(x − 12), and then (x − 12)(x + 8). This is the only place in the chapter where the grouping step is performed in full, and it is not given a name. It is what makes the leading-coefficient cases of Q1 workable.
• Exercise Set 4.4 Q2 (pp. 81–82), eight products to obtain by choosing an identity rather than multiplying: 41², 27², 23 × 17, 135², 97², 18 × 29, 34 × 43, 205². Verified: 41² = (40 + 1)² = 1681; 27² = (30 − 3)² = 900 − 180 + 9 = 729; 23 × 17 = (20 + 3)(20 − 3) = 400 − 9 = 391; 135² = (130 + 5)² = 16900 + 1300 + 25 = 18225, or by the p. 77 method as 140 × 130 + 25; 97² = (100 − 3)² = 10000 − 600 + 9 = 9409; 205² = (200 + 5)² = 42025. The two interesting ones are 18 × 29 and 34 × 43, because they are not symmetric about a whole number and so are not a difference of squares in integers. They are cases of this topic's own identity: 18 × 29 = (20 − 2)(20 + 9) = 400 + 20(−2 + 9) + (−2)(9) = 400 + 140 − 18 = 522, and 34 × 43 = (40 − 6)(40 + 3) = 1600 + 40(−6 + 3) + (−6)(3) = 1600 − 120 − 18 = 1462. Sequencing 23 × 17 (where the two parts cancel) before 18 × 29 (where they do not) is the cleanest way to show that a² − b² is the special case a + b = 0 of the wider identity.
• Exercise Set 4.4 Q3 (iii) (p. 82). Factorise r² − r − 42. Verified: (r − 7)(r + 6), since −7 + 6 = −1 and (−7)(6) = −42. Here the constant is negative, so the two parts have opposite signs and the larger magnitude carries the sign of the middle term — the third and last of the three sign patterns.
• Example 18, the pool (§4.8, p. 88). A rectangular pool whose breadth is 4 metres less than its length has area 96 square metres. Inputs: 4 and 96. The page sets the length to x, writes x(x − 4) = 96, rearranges to x² − 4x − 96 = 0, splits −4x, factorises to (x − 12)(x + 8) = 0, and then rejects the negative root. Verified: the split needs two numbers with product −96 and sum −4, namely −12 and 8; the roots are 12 and −8; length 12 m and breadth 8 m, and 12 × 8 = 96 checks out. A misprint to correct silently: the page states that (−12) × 8 = 96. It is −96. The stated sum on the same line, −4, is right, and the rest of the example is right; only that one product is wrong.
• When both roots survive (End-of-Chapter Q8, p. 90). A number plus its reciprocal is 10/3. Verified: 3x² − 10x + 3 = 0, which factorises as (3x − 1)(x − 3), so x = 3 or x = 1/3 — and both are genuine answers, because each is the other's reciprocal. Set beside the pool this makes the point that discarding a root is a decision about the situation, not a step in the method.
• A stronger conclusion than the question asks (End-of-Chapter Q10, p. 90, starred). Given that x − 2 and x − 1/2 are both factors of px² + 5x + r, show p = r. Verified: the product of the roots is r/p, and 2 × (1/2) = 1, so r = p directly. Worth noting: the data actually fixes the values, not just their equality — the sum of the roots is −5/p = 5/2, so p = −2 and r = −2, and −2x² + 5x − 2 = −(2x − 1)(x − 2). The question asks only for the weaker statement.

Figures to have open

• No figure from the chapter is required; §4.6 prints none. That absence is itself the topic's opening beat, so show the tile rectangle from Algebra tiles: factorising by rebuilding the rectangle once and then remove it.
• A factor-pair table with a sums column. Standard schematic, and the single most useful visual in this topic.
• A sign-pattern map for the three cases. Standard schematic, not in the book — the chapter states no such rule.
• A scale drawing of the 12 m by 8 m pool, and a rejected drawing with a negative side, so the discarded root has something to fail against. Standard schematic.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 4, "Exploring Algebraic Identities": §4.6 "Factorisation Without Using Algebra Tiles", pp. 80–81, Examples 10, 11 and 12.
• Exercise Set 4.4, pp. 81–82, Q1, Q2 and Q3 (iii).
• Example 18, printed in §4.8 on p. 88 — the only fully worked split-and-group in the chapter.
• End-of-Chapter Exercises Q8, Q9, Q10, p. 90.
• Chapter summary, p. 90, fourth bullet; the identity list on p. 91 carries both x² + (a + b)x + ab = (x + a)(x + b) and acx² + (ad + bc)x + bd = (ax + b)(cx + d).
• Companion topic: Algebra tiles: factorising by rebuilding the rectangle, which supplies the reason the two conditions come as a pair.