Chapter 5.3 — Exercise 5.2 — Factorisation Method
Solving quadratic equations using factorisation method. This is Lesson 3 of 5 in Chapter 5: Quadratic Equations.
Splitting a Quadratic Into Two Linear Pieces
A root of ax² + bx + c = 0 is any real number α satisfying aα² + bα + c = 0 exactly. Since a quadratic polynomial can have at most two zeroes, every quadratic equation has at most two roots — and factorisation finds both of them at once, by splitting the equation into two linear factors and setting each one to zero in turn.
The Splitting-the-Middle-Term Method
For 6x² − x − 2 = 0 (a=6, b=−1, c=−2), the method looks for two numbers p and q whose sum equals b and whose product equals a×c:
p + q = −1 p × q = 6 × (−2) = −12Checking the factor pairs of −12 one at a time — (1,−12), (−1,12), (2,−6), (−2,6), (3,−4), (−3,4) — only the pair p=3, q=−4 satisfies both conditions simultaneously. Splitting the middle term −x into the two pieces +3x and −4x:
6x²+3x−4x−2=0 ⟹ 3x(2x+1)−2(2x+1)=0 ⟹ (2x+1)(3x−2)=0Setting each factor to zero gives x = −1/2 or x = 2/3 — the equation's two roots. It's always worth checking at least one root back in the original equation before moving on: 6(−1/2)² − (−1/2) − 2 = 6(1/4) + 1/2 − 2 = 3/2 + 1/2 − 2 = 0, confirming the whole factorisation was carried out correctly.
Nine Equations, Factorised
The same p+q / p×q search, run systematically against nine different equations, produces every root needed anywhere in this exercise:
| Equation | Split | Roots |
|---|---|---|
| x² − 3x − 10 = 0 | p=−5, q=2 | 5, −2 |
| 2x² + x − 6 = 0 | p=4, q=−3 | −2, 3/2 |
| √2x² + 7x + 5√2 = 0 | p=2, q=5 | −√2, −5/√2 |
| 2x² − x + 1/8 = 0 (×16) | 16x²−8x+1=0, p=q=−4 | 1/4, 1/4 (repeated) |
| 100x² − 20x + 1 = 0 | p=q=−10 | 1/10, 1/10 (repeated) |
| x(x+4) = 12 | x²+4x−12=0, p=6, q=−2 | −6, 2 |
| 3x² − 5x + 2 = 0 | p=−3, q=−2 | 1, 2/3 |
| x − 3/x = 2 | x²−2x−3=0, p=−3, q=1 | 3, −1 |
| 3(x−4)² − 5(x−4) − 12 = 0 | substitute a=x−4: 3a²−5a−12=0 | 7, 8/3 |
Two rows produce a repeated root — the same value twice, not two different numbers — which happens exactly when the two factors turn out identical, as in 100x² − 20x + 1 = 0 splitting to (10x−1)(10x−1) = 0. The final row needs one extra move before splitting can even start: substituting a = x−4 turns a messy expression in x into a clean quadratic in a, which gets solved first and only converted back to x at the very end — a=3 becomes x=7, and a=−4/3 becomes x=−4/3+4=8/3.
The third row is worth a second look, since it's the only one where every coefficient involves √2. Multiplying a×c here gives (√2)(5√2) = 5×2 = 10, a whole number despite both original factors carrying a radical — which is exactly why p=2 and q=5 work as ordinary integers even though the equation itself is full of square roots. Splitting the middle term the same way as any other row, then factoring out √2 from the first pair and a plain constant from the second, produces the two roots directly.
Two Numbers From a Sum and a Product
Two numbers add to 27 and multiply to 182. Naming one x, the other is 27−x:
x(27−x) = 182 ⟹ x² − 27x + 182 = 0 ⟹ (x−14)(x−13) = 0This gives x = 14 or x = 13 — and either choice names the same pair of numbers in a different order, so the two numbers are simply 13 and 14. Unlike most word problems in this exercise, neither root here needs to be rejected — both describe the identical answer, just with the labels swapped.
Five Geometry Problems
Translating a shape's dimensions into one variable, exactly as practised in Exercise 5.1, turns each of these five geometry problems into a quadratic ready to factorise:
| Problem | Equation | Answer |
|---|---|---|
| Two consecutive positive integers, sum of squares 613 | 2x²+2x−612=0 | 17 and 18 |
| Right triangle: altitude 7 cm less than base, hypotenuse 13 cm | x²−7x−60=0 | Base 12 cm, altitude 5 cm |
| Rectangle: perimeter 28 m, area 40 m² | x²−14x+40=0 | 10 m and 4 m |
| Triangle: base 4 cm longer than altitude, area 48 cm² | x²+4x−96=0 | Altitude 8 cm, base 12 cm |
| Cottage industry: cost per article is 3 more than twice the count, total ₹90 | 2x²+3x−90=0 | 6 articles at ₹15 each |
The right-triangle problem leans on the Pythagorean theorem to set up its equation (base² + altitude² = hypotenuse²) before factorisation ever begins — a reminder that "formation" often draws on facts from outside algebra entirely, geometry in this case, to produce the equation that factorisation then solves. The rectangle problem is worth a second look too: perimeter 28 m fixes length+breadth at 14 m before area even enters the picture, so naming the length x automatically makes the breadth 14−x, and the area condition x(14−x)=40 is what actually produces the quadratic — two separate given facts, perimeter and area, combining into one equation in a single variable.
Motion and a Class Fundraiser
Three more problems each produce two mathematically valid roots — but only one root fits the real situation being described, so checking every answer against the physical constraints of the problem is genuinely part of solving it correctly, not an optional afterthought tacked on at the end.
- Two trains from the same station, one 5 km/h faster, travelling west and north for 2 hours end up 50 km apart (Pythagorean setup again): x²+5x−300=0 gives x=15 or x=−20 — only 15 is a valid speed, so the trains travel at 15 km/h and 20 km/h.
- 60 students, boys and girls contributing money equal to the other group's headcount, totalling ₹1600: x²−60x+800=0 gives x=20 or x=40 — both are valid here, since the problem doesn't distinguish which group is larger.
- A motor boat 24 km upstream against a 3 km/h current, round trip in 6 hours: x²−8x−9=0 gives x=9 or x=−1 — only 9 km/h is a valid still-water speed.
The class-fundraiser problem is the outlier worth noticing: most word problems in this exercise reject one root as physically impossible, but here both 20 and 40 describe a genuine, consistent scenario — 20 boys and 40 girls, or 40 boys and 20 girls, produce the identical ₹1600 total either way, since the equation itself never distinguishes which group happens to be labelled x. Only the physical setup of a problem, never the algebra alone, determines whether a second root should be kept or thrown away.
When Factorisation Gets Difficult
Every single equation solved here happened to split into two clean integer or simple-fraction factors. Not every quadratic equation cooperates that way, however carefully you search for a splitting pair. Exercise 5.3 introduces a method — completing the square — that works even when no such clean split exists, and derives the single general quadratic formula as a direct algebraic result of that process. For the root and degree definitions this exercise builds on, see the chapter introduction, or revisit the word-problem translation skill this exercise's own geometry and rate problems lean on, first practised in Exercise 5.1.