Sec 3 G1 Maths Tutor Bukit Timah

Did you know? The expression x² + 7x + 12 can be written as (x + 3)(x + 4) because 3 + 4 = 7 and 3 × 4 = 12. The same pair of numbers performs two different jobs: its sum produces the x coefficient and its product produces the constant.

This Secondary 3 G1 Mathematics Tutor Bukit Timah guide teaches quadratic factorisation of x² + px + q by finding integer factor pairs. Students learn what the coefficient and constant mean, how positive and negative pairs work, how to expand a proposed answer to check it, and how to recognise when an expression does not split into integer linear factors.

The SEAB 2027 G1 syllabus directory identifies Mathematics as K110. The official K110 Mathematics syllabus specifically includes factorisation of quadratic expressions of the form x² + px + q. It excludes treating the special identities (a ± b)² = a² ± 2ab + b² and a² − b² = (a + b)(a − b) as required factorisation techniques. This guide therefore uses sum-and-product matching and ordinary distribution; it does not introduce excluded identities as compulsory G1 shortcuts.

Quick answer: how does the sum-and-product method work?

For a quadratic of the form x² + px + q, find two integers m and n whose sum equals p and whose product equals q. Then write (x + m)(x + n). Always check by multiplying the two brackets back out: the x² coefficient should be one, the middle coefficient should be m + n, and the constant should be mn.

A good tutor asks for the check even when the factors look obvious. A factor pair that multiplies to the constant but has the wrong sum will produce a different quadratic. The verification step catches that before the error becomes a habit.

Three sign rules that simplify the search

Positive constant q: an integer factor pair has the same signs; the sign of the middle coefficient helps select positive or negative.
Negative constant q: the two integer factors have opposite signs.
Middle coefficient p: this is the sum of the signed factors, not their product.
Check: expand (x + m)(x + n) = x² + (m + n)x + mn by distribution and collecting like terms.

This is a search for integer factors, not a guarantee that every quadratic will have them. Some quadratics cannot be factorised into two brackets with integer constants. That is not evidence the student’s arithmetic failed; it may be a genuine feature of the coefficients. Read the school’s precise question and do not force an invalid factor pair.

Worked example: why x² + 7x + 12 uses 3 and 4

Start with the constant 12. The positive factor pairs include 1 and 12, 2 and 6, and 3 and 4. Their sums are 13, 8 and 7 respectively. The target middle coefficient is seven, so choose 3 and 4. Expanding (x + 3)(x + 4) gives x² + 4x + 3x + 12 = x² + 7x + 12. Both conditions agree.

This is different from solving x² + 7x + 12 = 0. Factorisation changes the expression into a product. An equation introduces the additional condition that the product is zero; it may then be solved using an appropriate method. Keep the task verbs ‘factorise’, ‘expand’ and ‘solve’ separate.

Twenty-eight original factor-pair worked examples

Factor-pair practice 1: x² + 7x + 12

Question. Factorise x² + 7x + 12 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 3 and n = 4.
Check the sum: 3 + (4) = 7, matching the middle coefficient.
Check the product: (3) × (4) = 12, matching the constant.
Write the factors: (x + 3)(x + 4).

Why this example matters: The target positive sum is found by testing a small pair of factors of twelve. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The pair 2 and 6 multiplies to twelve but sums to eight, not seven. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: Replace the middle coefficient with eight; then (x + 2)(x + 6) gives x² + 8x + 12. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 2: x² + 8x + 15

Question. Factorise x² + 8x + 15 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 3 and n = 5.
Check the sum: 3 + (5) = 8, matching the middle coefficient.
Check the product: (3) × (5) = 15, matching the constant.
Write the factors: (x + 3)(x + 5).

Why this example matters: Both the constant and middle coefficient are positive, so a pair of positive integers is a sensible starting point. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The pair 1 and 15 gives the right product but the sum sixteen is wrong. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: Changing q to sixteen while keeping p eight leads to (x + 4)(x + 4). Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 3: x² + 5x + 6

Question. Factorise x² + 5x + 6 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 2 and n = 3.
Check the sum: 2 + (3) = 5, matching the middle coefficient.
Check the product: (2) × (3) = 6, matching the constant.
Write the factors: (x + 2)(x + 3).

Why this example matters: A small factor pair makes the sum easy to check independently of the product. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Choosing 1 and 6 multiplies to six but produces a middle coefficient seven. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 7x + 6, the correct pair is 1 and 6. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 4: x² + 9x + 20

Question. Factorise x² + 9x + 20 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 4 and n = 5.
Check the sum: 4 + (5) = 9, matching the middle coefficient.
Check the product: (4) × (5) = 20, matching the constant.
Write the factors: (x + 4)(x + 5).

Why this example matters: Two close positive factors often generate a middle coefficient slightly below their product. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The factor pair 2 and 10 gives twelve rather than the required sum nine. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: Compare with x² + 12x + 20 = (x + 2)(x + 10). Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 5: x² + 6x + 8

Question. Factorise x² + 6x + 8 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 2 and n = 4.
Check the sum: 2 + (4) = 6, matching the middle coefficient.
Check the product: (2) × (4) = 8, matching the constant.
Write the factors: (x + 2)(x + 4).

Why this example matters: Listing factor pairs helps more than trying random bracket expressions. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: If m and n are both positive, do not put a minus sign in one bracket; that would change the constant’s sign. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: The expression x² − 6x + 8 instead uses (x − 2)(x − 4). Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 6: x² + 13x + 40

Question. Factorise x² + 13x + 40 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 5 and n = 8.
Check the sum: 5 + (8) = 13, matching the middle coefficient.
Check the product: (5) × (8) = 40, matching the constant.
Write the factors: (x + 5)(x + 8).

Why this example matters: The product forty has several factor pairs; only one pair sums to thirteen. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Using 4 and 10 would produce fourteen as the middle coefficient. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 14x + 40, (x + 4)(x + 10) is correct. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 7: x² + 14x + 48

Question. Factorise x² + 14x + 48 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 6 and n = 8.
Check the sum: 6 + (8) = 14, matching the middle coefficient.
Check the product: (6) × (8) = 48, matching the constant.
Write the factors: (x + 6)(x + 8).

Why this example matters: Even coefficients and an even constant can be checked with easy mental multiplication. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The pair 4 and 12 gives sixteen rather than fourteen. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 16x + 48, use (x + 4)(x + 12). Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 8: x² + 11x + 28

Question. Factorise x² + 11x + 28 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 4 and n = 7.
Check the sum: 4 + (7) = 11, matching the middle coefficient.
Check the product: (4) × (7) = 28, matching the constant.
Write the factors: (x + 4)(x + 7).

Why this example matters: It helps to write both the factor product and their sum beside the proposed brackets. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The pair 2 and 14 sums to sixteen, so it cannot be the required pair. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 9x + 14, choose 2 and 7. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 9: x² − 7x + 12

Question. Factorise x² − 7x + 12 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -3 and n = -4.
Check the sum: -3 + (-4) = -7, matching the middle coefficient.
Check the product: (-3) × (-4) = 12, matching the constant.
Write the factors: (x − 3)(x − 4).

Why this example matters: A positive constant with a negative middle coefficient points to two negative factors. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Choosing +3 and +4 makes the middle coefficient +7 instead of −7. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² − 8x + 12, choose −2 and −6. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 10: x² − 9x + 20

Question. Factorise x² − 9x + 20 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -4 and n = -5.
Check the sum: -4 + (-5) = -9, matching the middle coefficient.
Check the product: (-4) × (-5) = 20, matching the constant.
Write the factors: (x − 4)(x − 5).

Why this example matters: The product of two negative numbers is positive, while their sum remains negative. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The mixed-sign pair −4 and +5 would create a negative product, not +20. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 9x + 20, reverse both signs to obtain positive factors. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 11: x² − 11x + 24

Question. Factorise x² − 11x + 24 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -3 and n = -8.
Check the sum: -3 + (-8) = -11, matching the middle coefficient.
Check the product: (-3) × (-8) = 24, matching the constant.
Write the factors: (x − 3)(x − 8).

Why this example matters: A pair of negative factors can share a positive product without cancelling their signed sum. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Using −4 and −6 would produce −10x, not −11x. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² − 10x + 24, use −4 and −6. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 12: x² − 13x + 40

Question. Factorise x² − 13x + 40 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -5 and n = -8.
Check the sum: -5 + (-8) = -13, matching the middle coefficient.
Check the product: (-5) × (-8) = 40, matching the constant.
Write the factors: (x − 5)(x − 8).

Why this example matters: Check product and sum as separate conditions before committing to an answer. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Writing (x − 5)(x + 8) would make the constant negative and middle coefficient positive three. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: Changing the middle term to −14x while keeping +40 would use −4 and −10. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 13: x² − 15x + 54

Question. Factorise x² − 15x + 54 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -6 and n = -9.
Check the sum: -6 + (-9) = -15, matching the middle coefficient.
Check the product: (-6) × (-9) = 54, matching the constant.
Write the factors: (x − 6)(x − 9).

Why this example matters: The factors must multiply to fifty-four but also combine to the large negative middle coefficient. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Using −3 and −18 produces −21x, which is not the stated middle term. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² − 21x + 54, use −3 and −18. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 14: x² − 6x + 8

Question. Factorise x² − 6x + 8 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -2 and n = -4.
Check the sum: -2 + (-4) = -6, matching the middle coefficient.
Check the product: (-2) × (-4) = 8, matching the constant.
Write the factors: (x − 2)(x − 4).

Why this example matters: A small negative pair is a useful bridge from positive-factor examples. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Writing (x − 2)(x + 4) changes the constant to −8. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² − 5x + 6, the factors are −2 and −3. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 15: x² + 1x − 12

Question. Factorise x² + 1x − 12 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 4 and n = -3.
Check the sum: 4 + (-3) = 1, matching the middle coefficient.
Check the product: (4) × (-3) = -12, matching the constant.
Write the factors: (x + 4)(x − 3).

Why this example matters: A negative constant requires opposite signs, and the net sum determines which has larger magnitude. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Using −4 and +3 produces −1 as the middle coefficient, not +1. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² − x − 12, use −4 and +3. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 16: x² + 5x − 24

Question. Factorise x² + 5x − 24 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 8 and n = -3.
Check the sum: 8 + (-3) = 5, matching the middle coefficient.
Check the product: (8) × (-3) = -24, matching the constant.
Write the factors: (x + 8)(x − 3).

Why this example matters: A positive middle coefficient means the positive factor has the greater magnitude in this opposite-sign pair. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Pair +6 and −4 has product −24 but sum two, not five. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 2x − 24, +6 and −4 would be correct. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 17: x² − 2x − 15

Question. Factorise x² − 2x − 15 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -5 and n = 3.
Check the sum: -5 + (3) = -2, matching the middle coefficient.
Check the product: (-5) × (3) = -15, matching the constant.
Write the factors: (x − 5)(x + 3).

Why this example matters: The negative factor has larger magnitude, so the signed sum is negative while the product is negative. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Choosing +5 and −3 gives positive two, the wrong middle coefficient. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 2x − 15, +5 and −3 works. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 18: x² − 4x − 21

Question. Factorise x² − 4x − 21 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -7 and n = 3.
Check the sum: -7 + (3) = -4, matching the middle coefficient.
Check the product: (-7) × (3) = -21, matching the constant.
Write the factors: (x − 7)(x + 3).

Why this example matters: Large factors with different signs can produce a small negative sum. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The pair −3 and +7 would make a positive middle coefficient four. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 4x − 21, choose +7 and −3. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 19: x² + 3x − 28

Question. Factorise x² + 3x − 28 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 7 and n = -4.
Check the sum: 7 + (-4) = 3, matching the middle coefficient.
Check the product: (7) × (-4) = -28, matching the constant.
Write the factors: (x + 7)(x − 4).

Why this example matters: A difference of three between factor magnitudes produces the required +3x middle term. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The pair +14 and −2 multiplies to −28 but sums to twelve. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 12x − 28, +14 and −2 works. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 20: x² − 5x − 14

Question. Factorise x² − 5x − 14 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -7 and n = 2.
Check the sum: -7 + (2) = -5, matching the middle coefficient.
Check the product: (-7) × (2) = -14, matching the constant.
Write the factors: (x − 7)(x + 2).

Why this example matters: The sign pattern is determined by a negative constant and a negative coefficient. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Using +7 and −2 gives +5x and still −14, which is a different quadratic. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 5x − 14, +7 and −2 is correct. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 21: x² + 2x − 35

Question. Factorise x² + 2x − 35 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 7 and n = -5.
Check the sum: 7 + (-5) = 2, matching the middle coefficient.
Check the product: (7) × (-5) = -35, matching the constant.
Write the factors: (x + 7)(x − 5).

Why this example matters: The positive factor exceeds the negative factor in magnitude by two. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: The pair +35 and −1 sums to 34 and fails the middle coefficient. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 34x − 35, +35 and −1 would fit. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 22: x² − 4x − 12

Question. Factorise x² − 4x − 12 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -6 and n = 2.
Check the sum: -6 + (2) = -4, matching the middle coefficient.
Check the product: (-6) × (2) = -12, matching the constant.
Write the factors: (x − 6)(x + 2).

Why this example matters: The integer pair combines a negative product and a negative sum. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Writing −2 and +6 reverses the signed sum and gives +4x. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 4x − 12, +6 and −2 works. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 23: x² − 1x − 6

Question. Factorise x² − 1x − 6 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -3 and n = 2.
Check the sum: -3 + (2) = -1, matching the middle coefficient.
Check the product: (-3) × (2) = -6, matching the constant.
Write the factors: (x − 3)(x + 2).

Why this example matters: An x coefficient of −1 means the signed factors nearly cancel, leaving one negative. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Choosing +3 and −2 would give +1x instead of −1x. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + x − 6, use +3 and −2. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 24: x² + 4x − 21

Question. Factorise x² + 4x − 21 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 7 and n = -3.
Check the sum: 7 + (-3) = 4, matching the middle coefficient.
Check the product: (7) × (-3) = -21, matching the constant.
Write the factors: (x + 7)(x − 3).

Why this example matters: The large positive factor and smaller negative factor combine to the positive middle coefficient. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Forgetting the minus inside the second bracket produces +21 as the constant. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² − 4x − 21, −7 and +3 works. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 25: x² + 12x + 36

Question. Factorise x² + 12x + 36 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 6 and n = 6.
Check the sum: 6 + (6) = 12, matching the middle coefficient.
Check the product: (6) × (6) = 36, matching the constant.
Write the factors: (x + 6)(x + 6).

Why this example matters: A repeated integer factor pair still satisfies the sum and product tests. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Do not require a special-square identity as a shortcut; the ordinary factor-pair method is sufficient. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 10x + 25, the repeated factor pair is +5 and +5. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 26: x² − 6x + 9

Question. Factorise x² − 6x + 9 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = -3 and n = -3.
Check the sum: -3 + (-3) = -6, matching the middle coefficient.
Check the product: (-3) × (-3) = 9, matching the constant.
Write the factors: (x − 3)(x − 3).

Why this example matters: Two equal negative factors have a positive product and a negative middle sum. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Using +3 twice would create +6x, which fails the coefficient sign. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² − 8x + 16, the repeated pair is −4 and −4. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 27: x² + 0x + 0

Question. Factorise x² + 0x + 0 as far as possible over the integers using the sum-and-product method.

Find the pair: choose m = 0 and n = 0.
Check the sum: 0 + (0) = 0, matching the middle coefficient.
Check the product: (0) × (0) = 0, matching the constant.
Write the factors: (x + 0)(x + 0).

Why this example matters: Zero factors give x × x, illustrating that the general pattern also accommodates a zero middle coefficient and zero constant. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Do not confuse the zero coefficient with an instruction that x itself must equal zero; this is factorising an expression. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: For x² + 5x + 0, one factor pair is 0 and 5, giving x(x + 5). Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Factor-pair practice 28: x² + 0x + 5

Question. Factorise x² + 0x + 5 as far as possible over the integers using the sum-and-product method.

Try to find integers m and n with m + n = 0 and mn = 5.
If m + n = 0, then n = −m, so mn = −m².
That product cannot equal the positive number 5.
Conclude there is no integer factor pair for this requested form.

Why this example matters: No integer pair has both product +5 and sum zero. Opposite integers sum to zero but their product is non-positive. If two integer factors are proposed, expanding their brackets is the most direct independent check of both the middle coefficient and constant.

Common trap: Do not manufacture (x + 5)(x − 5): it expands to x² − 25, a different expression. The first correction should inspect which of the two conditions—sum or product—failed, not simply replace the final brackets.

Try a different version: The question ‘factorise over integers’ has no two-integer-bracket answer for x² + 5. Explain the reason instead of guessing. Do the new problem with the earlier answer hidden and write both a sum check and product check before completing the factorisation.

Expanding a pair of factors provides the definitive check

Suppose the proposed factorisation is (x + 8)(x − 3). Multiply every term in the first bracket by every term in the second: x² − 3x + 8x − 24. Combine the two x terms to obtain x² + 5x − 24. If the original expression was x² + 5x − 24, the check passes. If the original was x² + 3x − 24, the check fails even though the constant is correct.

This kind of error often arises when a learner finds an attractive pair of factors and stops after verifying the product. Two conditions are needed. Writing the expansion also makes the method available when the student forgets the shortcut. It is a way to establish algebraic equivalence, not just an exam presentation flourish.

What if no integer factor pair exists?

The task ‘factorise x² + px + q’ in school contexts generally uses examples chosen to have accessible integer factor pairs when the intended response is two integer brackets. But not every quadratic has that property. For x² + 5, integers m and n with sum zero would have n = −m, so product −m², which cannot be positive five. A student who cannot find a factor pair here is encountering a genuine limitation of this method.

It is better to state the limitation accurately than to present a false factorisation. The wider syllabus includes solving quadratic equations by the quadratic formula, which is a different task and works more broadly under appropriate conditions. It does not mean the excluded special-square identities need to be taught as mandatory shortcuts.

Factorising an expression versus solving an equation

Factorising x² + 7x + 12 produces (x + 3)(x + 4), an expression equal to the original for every x. If the question instead states x² + 7x + 12 = 0, the equivalent equation is (x + 3)(x + 4) = 0. For real-number factors, a product is zero when at least one factor is zero, giving x = −3 or x = −4.

That additional step uses the equality to zero; it does not belong in an ordinary ‘factorise’ question. Parents can ask children to read the command word at the start of a question: factorise, solve and expand require different final answer formats.

Negative signs: how to choose the right pair efficiently

For a positive constant, both factors have the same sign. If the x coefficient is negative, both are negative. For a negative constant, one factor is positive and one negative. The sign and size of the sum tells you which factor has the larger absolute value. This is a reasoning method, not a memorised list of unrelated cases.

For example, x² − 5x − 14 requires product −14. Candidate pairs are +7 and −2, whose sum is +5, and −7 and +2, whose sum is −5. The second pair matches. Expanding (x − 7)(x + 2) confirms x² − 5x − 14. One extra sign check avoids an entire wrong working line.

A tutor’s correction record for quadratic factorisation

The learner checks the product but not the sum

Give x² + 7x + 12. The pair +2 and +6 gives product twelve, but its sum eight is wrong. Require both comparisons on paper. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The learner forgets that factors can be negative

Compare x² + 7x + 12 with x² − 7x + 12. The constant stays positive while both factor signs reverse. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

A negative constant leads to two negative factors

A product of two negatives is positive. For q negative, select a positive and a negative factor, then verify their signed sum. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The student puts the large magnitude on the wrong sign

Compare pairs +7, −2 and −7, +2. Both give product −14 but their sums are +5 and −5, so the middle coefficient decides the choice. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The student confuses factorisation with solving

An expression does not supply x = 0 or any other value. Bracket factors are the answer when the instruction is ‘factorise’. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The student introduces an excluded identity as essential

The 2027 K110 syllabus excludes using the special-square and difference-of-squares identities as required techniques. Use ordinary factor-pair matching and expansion instead. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The learner treats every quadratic as integer-factorable

Some quadratics have no such integer pair. Verify whether the requested factorisation form is possible before inventing brackets. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The check is carried out with careless expansion

Write x × x, x × n, m × x and m × n, then combine the middle terms. A correct proposed factorisation must reconstruct the original quadratic. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The student loses the coefficient one

The method here targets monic x² + px + q. Expressions with another x² coefficient need other approaches not automatically required by this specific worked pattern. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

The learner relies on the look of factor numbers

A plausible pair is not evidence by itself. An explicit sum and product calculation is short, checkable and reliable. For independent practice, change only the middle coefficient or constant and ask the student to decide whether the original pair still works. The answer should include a reason, not just a guess at two new brackets.

A four-week G1 tutoring sequence for this skill

Week 1: factor pairs as arithmetic

Start with whole numbers such as twelve, fifteen, twenty and twenty-four. Ask for their positive factor pairs and the corresponding sums. Keep x out of the exercise initially. When students can list factors without missing a pair, introduce x² + 7x + 12 and explain how the target middle coefficient selects one pair.

Week 2: negative coefficients and opposite signs

Teach x² − 7x + 12 followed by x² + x − 12. Have learners predict the factor signs from the constant before doing detailed arithmetic. Compare a pair with its reversed signs and explain how the signed sum changes. A short number-line check may help students who still struggle with negative additions.

Week 3: expand every answer to verify

Use four medium examples and require complete expansion back to x² + px + q. Mix repeated pairs, positive constants and negative constants. If a pair fails, ask which condition failed rather than erasing the answer. The aim is to build the habit of checking algebraic equivalence.

Week 4: command words and problem transfer

Give a mixture of ‘expand’, ‘factorise’ and simple ‘solve’ questions within the school’s current scope. Include one expression that does not factorise into integer linear brackets so the learner must explain why the search does not succeed. The final check should be an unseen problem solved without a model solution.

Parents’ frequently asked questions

Is x² + px + q factorisation in the 2027 G1 syllabus?

Yes. The 2027 K110 Algebraic Expressions and Formulae strand specifies factorisation of quadratic expressions of this exact monic form. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

Does a positive constant mean both factors are positive?

Not always. Both may be negative too; the sign of the middle coefficient distinguishes the cases. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

Does a negative constant mean one factor is negative?

Yes, when the factor constants are real nonzero numbers whose product is negative. For an integer factor-pair method, one is positive and the other negative. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

How does my child know which two factors to choose?

Find pairs whose product is the constant, then identify the pair whose sum is the x coefficient. Both conditions must hold. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

Why are special square identities excluded?

That exclusion is stated in the official 2027 K110 syllabus. This article uses ordinary distribution and integer factor pairs rather than treating those identities as required techniques. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

Can a factorisation be checked with a calculator?

A calculator can help check arithmetic, but the most reliable algebraic verification is to expand the proposed brackets and compare coefficients. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

Is factorising the same as solving a quadratic?

No. Factorising rewrites an expression. Solving an equation determines values of the unknown that satisfy the equality, using an appropriate method. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

What if no integer pair seems to work?

Recheck all factor pairs and signs, then consider whether the expression actually has an integer factorisation. Not every quadratic does. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

How can tutoring show the learner has improved?

Give a new quadratic with a different sign pattern and ask for the factor pair, both checks and the expansion. Transfer is stronger evidence than repeating one example. A parent can change a single coefficient in the exercise and ask what changes in the two factors. This builds a habit of mathematical checking rather than reliance on one remembered pattern.

The ten-point factorisation checklist

Confirm the task is to factorise, not solve an equation.
Read the x² coefficient and check whether the target is the monic form x² + px + q.
Identify the middle coefficient p, including its sign.
Identify the constant q, including its sign.
List possible integer pairs whose product is q.
Use the required signed sum p to select a candidate.
Write the two monic brackets in the form (x + m)(x + n).
Expand and combine the middle terms to verify the expression.
Do not invent a factor pair where the integer conditions cannot be satisfied.
Give the final factorised expression rather than an unsupported value for x.

Read the connected Secondary 3 G1 guides

A Bukit Timah tutoring next step

Bring a recent polynomial question showing the student’s factor-pair choice and expansion check. The unchanged eduKateSG Secondary 1 Mathematics Tutor Clementi small-group tutorial reference describes close attention and guided corrections near Sixth Avenue MRT, but it is a teaching reference, not confirmation of a current Secondary 3 G1 vacancy or lesson timetable. Families can enquire about appropriate support.

Secondary 3 G1 Mathematics Tutor Bukit Timah | Quadratic Factorisation by Sum and Product