A quadratic expression is not yet a quadratic equation. Factorising is not yet solving. The equation becomes solvable by factorisation when the relationship is rewritten so that a product equals zero. Then the zero-product property turns one quadratic condition into simpler linear conditions.
This Secondary 2 Mathematics Learning Guide connects factorisation, roots, graph intercepts and word problems. The aim is not to memorise a sequence of moves. It is to understand why the moves are valid, when they are useful, and how to reject algebraic answers that violate the original context.
Secondary Mathematics Hub: S1–S4 Capability Map · Secondary 2 Learning Guide, Batch 2, Guide 4. Companion guides cover quadratic functions and graphs, linear inequalities, and simultaneous linear equations.
Course boundary. The current MOE G2/G3 Mathematics syllabus includes Secondary Two solving quadratic equations in one variable by factorisation. This guide stays with factorisation-based solving and related modelling. Methods such as the quadratic formula, discriminant analysis or completing the square belong to later or different courses unless the school introduces them separately.
Navigate: Expression versus equation · Zero-product property · Factorise then solve · Rearrange to zero · Roots and graphs · Problem solving · Practice and answers · Teaching and transfer.
1. Start by identifying the mathematical object
The expression x² + 5x + 6 can be expanded, factorised or evaluated, but it cannot be solved because no equality has been stated. The equation x² + 5x + 6 = 0 asks which values of x make the expression equal to zero.
The same expression can therefore appear in different jobs. Factorise x² + 5x + 6 gives (x + 2)(x + 3). Solve x² + 5x + 6 = 0 gives x = −2 or x = −3. The factorised form is a step toward the solution, not the solution itself.
Worked example 1: distinguish three tasks
Expression: x² − 7x + 10. Factorise: (x − 5)(x − 2). Solve x² − 7x + 10 = 0: x = 5 or x = 2. Evaluate at x = 3: 9 − 21 + 10 = −2.
These answers are not interchangeable. A learner who writes x = 5 or 2 for a factorisation question has performed an extra operation the question did not ask for. A learner who stops at the factorised form in a solving question has stopped too early.
2. The zero-product property is the bridge from factors to roots
If two real numbers multiply to zero, at least one factor must be zero. Therefore, if (x − 4)(x + 7) = 0, either x − 4 = 0 or x + 7 = 0. The roots are x = 4 and x = −7.
This property is special to a product equal to zero. If (x − 4)(x + 7) = 12, you cannot set either factor to zero. The equation must first be transformed into an equivalent form with zero on one side before factorisation can lead to roots.
Worked example 2: solve from an already-factorised form
Solve (2x + 3)(x − 5) = 0. Set each factor equal to zero. From 2x + 3 = 0, x = −3/2. From x − 5 = 0, x = 5.
The two roots are x = −3/2 or x = 5. Fractional roots are valid. The structure of the factors, not a preference for integers, determines the solutions.
Do not divide away a possible root
For x(x − 6) = 0, the roots are x = 0 and x = 6. Dividing both sides by x would produce x − 6 = 0 and lose x = 0 because division by x assumes x is non-zero.
The safer route is to use the zero-product property directly. If a later method involves division by an expression containing the unknown, first consider whether that expression can equal zero.
3. Factorise the quadratic completely before solving
For x² + 9x + 20 = 0, seek two numbers with product 20 and sum 9. The pair 4 and 5 gives (x + 4)(x + 5) = 0. Therefore x = −4 or x = −5.
When the constant is negative, the two factor numbers have opposite signs. For x² + x − 12 = 0, the pair 4 and −3 has product −12 and sum 1, so (x + 4)(x − 3) = 0. The roots are −4 and 3.
Worked example 3: signs carry the middle term
Solve x² − 11x + 24 = 0. The product is +24 and the sum is −11, so both factor numbers are negative: −3 and −8. Therefore (x − 3)(x − 8) = 0.
The solutions are x = 3 or x = 8. Check the middle term by expanding: −8x − 3x = −11x. This expansion check catches many sign mistakes before they become wrong roots.
Worked example 4: leading coefficient not equal to 1
Solve 2x² + 7x + 3 = 0. Factorise as (2x + 1)(x + 3) = 0 because expansion gives 2x² + 6x + x + 3.
Hence 2x + 1 = 0 or x + 3 = 0, giving x = −1/2 or x = −3. Do not apply the simple pair-sum rule as though the leading coefficient were 1.
Worked example 5: common factor before quadratic factors
Solve 3x² − 15x = 0. Factorise the common factor first: 3x(x − 5) = 0. Since 3 is non-zero, the zero product comes from x = 0 or x − 5 = 0. Thus x = 0 or x = 5.
Dividing the original equation by x would again lose x = 0. Factorisation keeps every candidate visible.
4. Move everything to one side before using the zero-product property
Quadratic equations often arrive with non-zero expressions on both sides. The zero-product method becomes available after the equation is rearranged so that one side is zero.
Worked example 6: x² equals a linear expression
Solve x² = 5x + 14. Move everything to the left: x² − 5x − 14 = 0. Factorise: (x − 7)(x + 2) = 0.
Therefore x = 7 or x = −2. Check in the original equation: 49 = 35 + 14 and 4 = −10 + 14. Both work.
Worked example 7: avoid setting factors to an unrelated number
Solve x(x + 4) = 12. It is not valid to write x = 12 or x + 4 = 12. Instead expand and rearrange: x² + 4x − 12 = 0.
Factorise as (x + 6)(x − 2) = 0, giving x = −6 or x = 2. Substitution into the original confirms both: −6(−2) = 12 and 2(6) = 12.
Equivalent rearrangements should lead to the same roots
From x² = 5x + 14, moving everything right instead gives 0 = −x² + 5x + 14. Multiplying by −1 gives 0 = x² − 5x − 14. These are equivalent equations with the same roots.
The route may differ, but valid transformations preserve the solution set. If two methods produce different roots, at least one transformation or factorisation has failed.
5. Roots are the x-intercepts of the related quadratic graph
The equation x² − 5x − 14 = 0 asks where the function y = x² − 5x − 14 has y-coordinate zero. These are the x-intercepts of its graph. Factorisation gives roots 7 and −2, so the graph crosses the x-axis at (7, 0) and (−2, 0).
This connects directly to Quadratic Functions, Graphs and Turning Points. The graph makes the roots visible spatially; the equation states the same condition symbolically.
Worked example 8: repeated root
Solve x² − 6x + 9 = 0. Factorise as (x − 3)² = 0. The only root is x = 3, repeated in the factorisation.
The related graph y = (x − 3)² touches the x-axis at (3, 0) and turns there rather than crossing. The algebraic repetition and graphical touching describe the same structure.
Not every quadratic factorises over the integers
The expression x² + x + 1 has no integer factor pair that multiplies to 1 and adds to 1. In a Secondary 2 factorisation-solving exercise, do not invent factors simply because every preceding example factorised neatly.
If the school task explicitly expects factorisation, first recheck the copied equation and any common factors. The broader mathematical world contains other solving methods, but this guide does not assume them.
6. Word problems create equations; context filters the roots
A modelling problem has two distinct stages. First convert the stated relationship into an equation. Then solve and interpret the roots. A negative algebraic root may be valid for the equation while invalid as a physical length or object count.
Worked example 9: rectangle dimensions
A rectangle has width x cm and length x + 4 cm. Its area is 96 cm². Find its dimensions. The area condition gives x(x + 4) = 96.
Rearrange: x² + 4x − 96 = 0. Factorise: (x + 12)(x − 8) = 0. The roots are x = −12 and x = 8. A physical width must be positive, so x = 8.
The dimensions are 8 cm by 12 cm. Check both conditions: the length is 4 cm greater than the width and 8 × 12 = 96.
Worked example 10: consecutive positive integers
The product of two consecutive positive integers is 156. Find them. Let the smaller integer be n. Then n(n + 1) = 156.
Rearrange: n² + n − 156 = 0. Factorise as (n + 13)(n − 12) = 0, giving n = −13 or n = 12. The positive-integer condition selects n = 12, so the integers are 12 and 13.
Worked example 11: a fixed-perimeter rectangle
A rectangle has perimeter 30 cm and area 54 cm². Find its side lengths. Let one side be x. Since 2(length + width) = 30, the other side is 15 − x.
Area gives x(15 − x) = 54. Rearranging gives x² − 15x + 54 = 0. Factorise: (x − 6)(x − 9) = 0, so x = 6 or 9. These roots simply swap which side is called x. The dimensions are 6 cm by 9 cm.
Two roots can represent the same physical rectangle
In the fixed-perimeter example, both 6 and 9 are positive and valid because choosing x = 6 makes the other side 9, while choosing x = 9 makes the other side 6. They describe the same unordered pair of side lengths.
This is different from the previous rectangle example where a negative root had to be rejected. Context does not always reduce two algebraic roots to one. It tells us how to interpret each candidate.
Worked example 12: a number problem with two valid roots
A number x satisfies x(x − 7) = 18. Find x. Rearrange: x² − 7x − 18 = 0. Factorise as (x − 9)(x + 2) = 0.
Therefore x = 9 or x = −2. Both satisfy the original equation. Unless the question adds a condition such as positive number, both answers must be retained.
7. Mixed practice: factorise, solve, interpret
Show the factorised equation before writing roots. For word problems, define the variable, formulate the equation, solve, interpret and check. Do not reject a root merely because it is negative unless the original meaning excludes it.
Questions 1–6. 1. Solve x² + 7x + 12 = 0. 2. Solve x² − x − 20 = 0. 3. Solve x² − 10x + 21 = 0. 4. Solve 2x² + 5x + 2 = 0. 5. Solve 4x² − 20x = 0. 6. Solve (3x − 2)(x + 5) = 0.
Questions 7–12. 7. Solve x² = 6x + 16. 8. Solve x(x + 3) = 40. 9. Solve x² − 8x + 16 = 0. 10. Find the x-intercepts of y = x² + 2x − 15. 11. A rectangle has width x and length x + 2, area 48. Find the dimensions. 12. The product of two consecutive positive integers is 210. Find them.
Questions 13–18. 13. A rectangle has perimeter 26 cm and area 40 cm². Find its side lengths. 14. Solve x(x − 4) = 21. 15. Explain why x(x − 4) = 21 cannot be solved by setting x = 0 or x − 4 = 0. 16. Explain why dividing x(x − 7) = 0 by x may lose a solution. 17. A quadratic equation gives roots −6 and 4, but x represents a positive length. Which root is admissible? 18. A quadratic equation gives roots 3 and 8, and x is an unrestricted real variable. How many answers should be reported?
Explained answers: questions 1–6
1. (x + 3)(x + 4) = 0, so x = −3 or −4. 2. (x − 5)(x + 4) = 0, so x = 5 or −4. 3. (x − 3)(x − 7) = 0, so x = 3 or 7.
4. (2x + 1)(x + 2) = 0, so x = −1/2 or −2. 5. 4x(x − 5) = 0, so x = 0 or 5. 6. x = 2/3 or −5.
Explained answers: questions 7–12
7. x² − 6x − 16 = 0 = (x − 8)(x + 2), so x = 8 or −2. 8. x² + 3x − 40 = 0 = (x + 8)(x − 5), so x = −8 or 5. 9. (x − 4)² = 0, so x = 4.
10. Factorise as (x + 5)(x − 3), so intercepts are (−5, 0) and (3, 0). 11. x(x + 2) = 48 gives (x + 8)(x − 6) = 0. The positive width is 6, so dimensions are 6 by 8. 12. n(n + 1) = 210 gives (n + 15)(n − 14) = 0, so the positive integers are 14 and 15.
Explained answers: questions 13–18
13. Let one side be x and the other 13 − x. Then x(13 − x) = 40, giving x² − 13x + 40 = 0 = (x − 5)(x − 8). The dimensions are 5 by 8. 14. x² − 4x − 21 = 0 = (x − 7)(x + 3), so x = 7 or −3.
15. The zero-product property applies only when the product equals zero; first rearrange to zero. 16. Division by x assumes x ≠ 0 and therefore removes x = 0 from consideration. 17. Only 4 is admissible as a positive length. 18. Both 3 and 8 should be reported.
8. Teach the chain: equation → zero → factors → roots → interpretation
A useful teaching sequence begins with already-factorised equations such as (x − 2)(x + 3) = 0 so the zero-product property is visible. Then move to factorisable quadratics already equal to zero. Only after those two stages are secure should students solve equations that must first be rearranged.
For foundation repair, use simple integer roots and make the learner expand the factors back as a check. For consolidation, mix unit and non-unit leading coefficients. For extension, use modelling questions in which two roots must be interpreted differently.
Diagnose the first false decision
If factorisation is wrong, repair factor structure. If factorisation is correct but the learner writes only one root, repair the zero-product property. If roots are correct but a negative length is retained, repair interpretation. If the learner sets factors to zero while the product equals 12, repair the condition for using the property.
These are distinct capabilities. Repeating twenty more quadratic equations may hide rather than isolate the weak link.
Questions parents can ask
Is the expression equal to zero yet? Have you factorised completely? Why are you allowed to set each factor to zero? How many roots should the equation have from this factorisation? Do both roots make sense in the original story? Can you substitute them back?
9. The transfer test: one equation, three representations
Consider x² − 6x + 8 = 0. The factorised form is (x − 2)(x − 4) = 0, giving roots 2 and 4. The related graph y = x² − 6x + 8 crosses the x-axis at (2, 0) and (4, 0). The symmetry line lies halfway between them at x = 3.
Now suppose the same equation arises from x(6 − x) = 8, where x represents a positive length and 6 − x is another positive length. Both x = 2 and x = 4 are admissible, and they simply exchange the two side lengths. The algebra has not changed, but the context tells us how the roots relate.
Use Algebraic Factorisation and Structural Control when factor construction is the weak link. Use Quadratic Functions, Graphs and Turning Points to connect roots with x-intercepts and symmetry. Use Simultaneous Linear Equations when a problem contains two independent linear constraints rather than one quadratic constraint.
Identify the equation. Rearrange to zero. Factorise completely. Use the zero-product property. Interpret every root in the original problem.
