Splitting the middle term, then setting each factor to zero

What to assume they know

• Root of an equation, zero of a polynomial, and the identification between them — Why a root of the equation is the same thing as a zero of the polynomial
• Standard form, and reading a, b and c with their signs — The shape an equation has to have before it counts as quadratic
• Factorising quadratic polynomials by splitting the middle term, from Class IX
• Taking a common factor out of a two-term expression, including a surd factor
• Solving a linear equation in one variable, including one with a fractional answer
• Arithmetic with √2, √3 and √6, and recognising √6 as √2 times √3

What they should be able to do

• State the property of zero that licenses equating each factor to zero, and explain why the same move fails when the product equals any other number
• Given ax² + bx + c, name the two conditions the split of the middle term must meet, and explain where those conditions come from
• Carry out the split, group in pairs, extract the common factor from each pair, and read off the two linear factors
• Convert a factorised equation into two linear equations and solve each
• Handle a split whose two numbers have opposite signs, and one whose numbers are surds
• Recognise when the two factors are identical and report the root the way the chapter does
• Return to a modelled situation, reject the root the situation cannot use, and state the answer in the units of the original problem
• Verify a claimed root by substitution before accepting it
• Say what factorising cannot do — supply a factorisation when no convenient one exists — and name the section of the chapter that answers that

Where it usually goes wrong

• "If the product equals any number, split it the same way." Only zero has the property being used. A product equal to 124 splits in unlimited ways, and the chapter's own Example 1 sets that trap by handing you exactly such a product.
• "Both factors must be zero." At least one must be. The two cases are taken in turn, which is why a quadratic yields two candidate roots rather than one simultaneous condition.
• "Any split of the middle term will do." The split must meet both conditions — the right sum and the right product — or the grouping produces two different brackets and nothing factors out.
• "The split numbers have to be whole numbers." Example 5 splits into two surds, and the exercise repeats the move with √2. The conditions are about arithmetic, not about integers.
• "A repeated factor means one root." The chapter reports the value twice, once for each occurrence of the factor, and §4.5 later calls this the coincident case.
• "The negative root is wrong." It is a perfectly good root of the equation. It is the situation that cannot use it, and saying so precisely is what Example 6 models.
• "Every quadratic can be factorised by inspection." Most cannot, which is the reason the chapter moves on to a formula. A student who believes otherwise will read §4.4 as an alternative rather than as a necessity.

Questions to check understanding

• Solving a supplied list of quadratics by factorising them — the standing instruction of Exercise 4.2 Question 1
• Items with a surd coefficient, where the split must be made in surds
• Items that factorise as a perfect square, where the expected answer names the root twice
• Word problems returning to a formed equation and requiring the inadmissible root to be identified and rejected with a reason
• Given one root, find the other, or find an unknown coefficient
• Verification items, where the marks are for the substitution rather than the root

Examples worth working on the board

Values marked verified are worked out here from the printed data; this chapter prints no answers, and every result below is derived here rather than looked up.

• Why zero and nothing else. Not printed, and needed early: if two numbers multiply to 12, that tells you almost nothing — 3 and 4, or 2 and 6, or −1 and −12, or 1/2 and 24 all qualify. If two real numbers multiply to nothing, at least one of them is nothing, with no alternatives at all. That asymmetry is the entire reason the chapter insists on standard form, with everything gathered on one side.
• The two conditions on the split. Given ax² + bx + c, look for two numbers whose sum is b and whose product is the product of a and c. The chapter states its splits with a bracketed justification of exactly this shape at Example 3, where it checks that the two pieces multiply back to the product of the outer terms. If the polynomial is (px + q)(rx + s), then the x term is (ps + qr)x while a is pr and c is qs, so ps and qr are two numbers with sum b and product ac.
• Example 3 (§4.3, p. 42). Equation 2x² − 5x + 3 = 0. Verified: the product of the outer coefficients is 6 and the middle coefficient is −5, so the split is −2 and −3; the four terms group as 2x(x − 1) − 3(x − 1), giving (2x − 3)(x − 1) = 0; the two linear equations yield 3/2 and 1. Substitution check at 3/2: 9/2 − 15/2 + 3 comes to nothing. At 1: 2 − 5 + 3 comes to nothing.
• Example 4 (§4.3, p. 43). Equation 6x² − x − 2 = 0. Verified: the outer product is −12 and the middle coefficient is −1, so the split is 3 and −4; the grouping is 3x(2x + 1) − 2(2x + 1), giving (3x − 2)(2x + 1) = 0, and the roots are 2/3 and −1/2. Substitution check at 2/3: 8/3 − 2/3 − 2 comes to nothing. At −1/2: 3/2 + 1/2 − 2 comes to nothing. This is the case where the two split numbers carry opposite signs, which is what a negative outer product always forces.
• Example 5 (§4.3, p. 43). Equation 3x² − 2√6x + 2 = 0. Verified: the outer product is 6 and the middle coefficient is −2√6, so the split is −√6 and −√6, since √6 times √6 is 6 and the two add to −2√6; the grouping gives √3x(√3x − √2) − √2(√3x − √2), so both factors are √3x − √2 and the equation is that factor squared. The single linear equation gives x = √2⁄√3, which the chapter reports twice, once for each occurrence of the factor.
• Example 6 (§4.3, p. 44), which closes the prayer hall of §4.1. Equation 2x² + x − 300 = 0, where x is the breadth in metres. Verified: the outer product is −600 and the middle coefficient is 1, so the split is −24 and 25; the grouping is 2x(x − 12) + 25(x − 12), giving (x − 12)(2x + 25) = 0, and the roots are 12 and −12.5. A breadth cannot be negative, so the breadth is 12 m and the length, being one more than twice it, is 25 m. Verified: 12 × 25 is 300, matching the stated carpet area.
• Exercise 4.2 (p. 44). Question 1 gives five equations to factorise: x² − 3x − 10 = 0; 2x² + x − 6 = 0; √2x² + 7x + 5√2 = 0; 2x² − x + 1/8 = 0; and 100x² − 20x + 1 = 0. Verified: the roots are 5 and −2; 3/2 and −2; −√2 and −5⁄√2; 1/4 taken twice; and 1/10 taken twice. The last two are the exercise's repeated-factor cases, and the third is its surd case — together they mirror Examples 3 to 5 exactly, so the exercise is a rerun rather than a step up. Question 2 sends the student back to Example 1. Verified: x² − 45x + 324 = 0 factorises with 9 and 36, and both roots are usable — one child began with 9 marbles and the other with 36, and the two roots are the same pair with the names swapped; x² − 55x + 750 = 0 factorises with 25 and 30, and both are usable, so the toy problem genuinely has two answers. Questions 3 to 6 are four more situations: two numbers with sum 27 and product 182; two consecutive positive integers whose squares total 365; a right triangle whose altitude is 7 cm short of its base and whose hypotenuse is 13 cm; and a pottery workshop whose per-article cost in rupees is 3 more than twice the day's output, with a total cost of ₹90. Verified: 13 and 14; 13 and 14 again; base 12 cm with altitude 5 cm, since 144 and 25 total 169; and 6 articles at ₹15 each.
• A trap worth staging. Example 1's marble situation reaches the product of two brackets equal to 124. Students who have just learned to equate factors to zero will try to equate them to 124. Verified: that would give x = 129 and x = −84, neither of which is a root of x² − 45x + 324 = 0. Show the failure, then gather to one side and factorise properly.

Figures to have open

• A split-and-group board: the quadratic on top, the four terms after splitting, the two pairs bracketed, the common factor drawn out of each. This is the workhorse of the topic and needs to be reusable across Examples 3 to 6. Standard schematic; §4.3 prints no figures.
• A "product equals zero" panel contrasting the many factor pairs of 12 with the single conclusion from a product of nothing. Standard schematic, an added device.
• Fig. 4.1 (p. 38) recalled in section 9, so the numbers 12 and 25 land back on the rectangle they came from. Redraw as a schematic.
• A number-line strip for section 9 showing 12 and −12.5, with the negative side shaded out as unavailable to a breadth. Standard schematic.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 4 "Quadratic Equations", §4.3 Solution of a Quadratic Equation by Factorisation, pp. 42–44 — the method, and Examples 3, 4, 5 and 6
• Same chapter, Exercise 4.2, p. 44 — Questions 1 to 6
• Same chapter, §4.1, p. 38 with Fig. 4.1 — the hall that Example 6 completes
• Same chapter, §4.2, Example 1, pp. 39–40 — the two situations Exercise 4.2 Question 2 sends the student back to
• Same chapter, §4.5 Summary, p. 47, point 3 — the restatement of the method
• Backward pointer: Class IX, for splitting the middle term as a factorising technique, which this chapter says it is reusing rather than teaching afresh