📈 Quadratics: factorise, complete the square, formula

Answer key

1. (c) 6x(2x + 3) — The highest common factor of 12x² and 18x is 6x. Dividing each term by 6x gives 12x² ÷ 6x = 2x and 18x ÷ 6x = 3, so 12x² + 18x = 6x(2x + 3). A candidate who only takes out the number 6 (missing the x) gets 6(2x² + 3x), which is not fully factorised. A candidate who only takes out 2x (missing the extra factor of 3 in 6) gets 2x(6x + 9), also not fully factorised — the bracket still shares a common factor. A candidate who takes out 3x instead of the full 6x gets 3x(4x + 6), which again is not fully factorised since 4x + 6 shares a common factor of 2.
2. (d) 12 m — Method: give the width a letter, write the length in terms of it and use length × width = area to form a quadratic. Working: with a width of x metres the length is x + 4, so x(x + 4) = 96, which rearranges to x² + 4x − 96 = 0; factorising gives (x + 12)(x − 8) = 0, and a width must be positive, so x = 8 and the length is 8 + 4 = 12. Answer: 12 m, and 12 × 8 = 96 as the area requires. The distractors: 8 m is the width rather than the length asked for; 16 m comes from taking the other root as 12 and adding 4 to it, ignoring that the root −12 cannot be a width; 24 m comes from dividing the area by the 4 in the question instead of forming an equation.
3. (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
4. (a) It touches the x-axis once, only at x = 4. — (x − 4)² is a square, so it equals zero only when x − 4 = 0, that is at x = 4 — the curve just touches the x-axis there rather than crossing it, since a square cannot be negative on either side to cross through. Saying it crosses at x = 4 and x = −4 wrongly introduces a plus-or-minus, as if taking a square root of x, rather than recognising the bracket is already squared and only zero once. Saying it never touches the x-axis forgets that a squared term CAN equal zero, even though it can never be negative. Saying it crosses at x = 2 and x = −2 confuses (x − 4)² with the different expression x² − 4.
5. (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
6. (b) (2x + 1)/(x − 3) — Factorise both the numerator and the denominator before you cancel anything. The numerator 2x² + 7x + 3 factorises to (2x + 1)(x + 3), and the denominator x² − 9 is a difference of two squares, factorising to (x − 3)(x + 3). The (x + 3) factor is common to both, so it cancels, leaving (2x + 1)/(x − 3). Writing (2x + 1)/(x + 3) comes from factorising x² − 9 as (x + 3)² instead of (x − 3)(x + 3) — a difference of two squares always has one plus and one minus bracket. Writing (2x + 3)/(x − 3) comes from mis-factorising the numerator as (2x + 3)(x + 1) and then wrongly cancelling the (x + 1) against the denominator's (x + 3) as though they were the same bracket. Writing 2x + 1 comes from cancelling the (x + 3) factor correctly but then dropping the remaining (x − 3) on the denominator altogether.
7. (b) x = −3, y = −7 — x² + 6x + 2 = (x + 3)² − 3² + 2 = (x + 3)² − 7. Substituting x = −3: (−3)² = 9, 6 × (−3) = −18, so 9 − 18 + 2 = −7, confirming the minimum value −7 at x = −3: turning point (−3, −7). Using 6 instead of half of 6 inside the bracket gives (x + 6)² − 34, turning point (−6, −34) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (3, −7) — wrong, because (x + 3)² is zero at x = −3, not x = 3. Computing 9 − 2 = 7 instead of 2 − 9 = −7 flips the sign of the constant, giving (−3, 7) — wrong, since the completed square's constant must be evaluated as 2 minus 9, not 9 minus 2. Always check a turning point by substituting its x-value back into the original equation.
8. (d) x = 2 or x = −5 — Method: find two numbers that multiply to give −10 and add to give 3 — these are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2. Distractor origins: x = −2 or x = 5 swaps the signs of the two roots; x = 2 or x = 5 makes both roots positive, ignoring the sign of −10; x = −5 or x = −2 makes both roots negative.
9. (d) (3x − 2)(x + 4) — For 3x² + 10x − 8, find two numbers multiplying to 3 × (−8) = −24 and adding to 10: these are 12 and −2. Rewrite: 3x² + 12x − 2x − 8 = 3x(x + 4) − 2(x + 4) = (3x − 2)(x + 4). Choosing (3x + 2)(x − 4) expands to 3x² − 10x − 8 — the correct factor pair but the wrong signs, giving the middle term the wrong sign. Choosing (x − 2)(3x + 4) expands to 3x² − 2x − 8 — the 3 is attached to the wrong bracket, changing which terms combine for the x-coefficient. Choosing (3x − 4)(x + 2) expands to 3x² + 2x − 8 — this uses 4 and 2 instead of the correct pair 12 and 2, so the middle term does not come to 10x.
10. (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
11. (c) x = 2, y = 5 — x² − 4x + 9 = (x − 2)² − 2² + 9 = (x − 2)² + 5. Substituting x = 2: 2² = 4, 4 × 2 = 8, so 4 − 8 + 9 = 5, confirming the minimum value 5 at x = 2: turning point (2, 5). Using −4 instead of half of it inside the bracket gives (x − 4)² − 7, turning point (4, −7) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−2, 5) — wrong, because (x − 2)² is zero at x = 2, not x = −2. Computing 4 − 9 = −5 instead of 9 − 4 = 5 flips the sign of the constant, giving (2, −5) — wrong, since the completed square's constant must be evaluated as 9 minus 4, not 4 minus 9. Substitute the x-value back into the original equation whenever you are unsure of a sign.
12. (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
13. (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
14. (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
15. (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
16. (d) x = 4 or x = −4 — Method: divide both sides by 2 to get x² = 16, then take the square root of both sides: x = 4 or x = −4. Distractor origins: x = 4 forgets the negative root; x = 8 comes from halving 16 instead of taking its square root; x = 16 stops at x² = 16 without ever taking the square root.
17. (d) 2x/(x − 2) — Factorise x² − 4 as (x − 2)(x + 2) first — it's a difference of two squares. The (x + 2) in the numerator then cancels with the (x + 2) in the factorised denominator, and 6x² ÷ 3x simplifies to 2x, leaving 2x/(x − 2). Writing 2/(x − 2) comes from over-cancelling 6x² ÷ 3x as 2 instead of 2x, dropping the x that should remain. Writing 2x/(x + 2) comes from factorising x² − 4 as (x + 2)² instead of (x − 2)(x + 2), a difference of two squares always has one plus and one minus bracket. Writing 2x²/(x − 2) comes from simplifying 6x² ÷ 3x as 2x² instead of 2x, dividing the coefficients but not reducing the power of x.
18. (c) 4 — x² + 12x + 40 = (x + 6)² − 6² + 40 = (x + 6)² + 4, so a = 6 and b = 4. Since (x + 6)² can never be negative, y = (x + 6)² + 4 is smallest when the bracket is zero, so the minimum value of y is 4. Giving 40 instead reads off the ORIGINAL constant term and ignores the completing-the-square step entirely — wrong, because 40 is the value of y when x = 0, not the minimum value. Giving −6 instead answers with the x-coordinate of the turning point (where the bracket is zero) rather than the minimum y-value itself — wrong, because the question asks for the minimum value of y, not the value of x that produces it. Giving 36 instead stops after squaring half the coefficient, 6² = 36, without combining it with the 40 already in the expression — wrong, because the minimum value is 40 minus 36, not 36 on its own.
19. (d) If the graph has two real roots, they are equal and opposite in value, so they sum to zero. — A turning point on the y-axis means the graph's axis of symmetry is the line x = 0, so any two roots must be symmetrical about x = 0 — equal in size but opposite in sign, summing to zero. The graph could still have no real roots, but that is not guaranteed just from the turning point's position, so the option claiming it must have none is too strong. The roots do not have to both be positive — if real, one is positive and one negative (or both are zero). The graph does not have to touch the x-axis at exactly one point either; it could cross at two symmetrical points, touch at one point, or miss the x-axis entirely.
20. (d) x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3).
21. (b) k = 9 — Method: complete the square, because a squared bracket is equal to zero for exactly one value of x. Working: x² − 6x = (x − 3)² − 9, so the equation becomes (x − 3)² − 9 + k = 0, that is (x − 3)² = 9 − k; there is exactly one solution when 9 − k = 0, so k = 9 and the equation reads x² − 6x + 9 = 0 with the repeated solution x = 3. Answer: k = 9. The distractors: k = 3 comes from halving the 6 and not squaring the result; k = 36 comes from squaring the whole of 6 instead of half of it; k = −9 comes from solving 9 − k = 0 with the sign of k the wrong way round.
22. (d) x³ + 6x² + 11x + 6 — (x + 1)(x + 2) = x² + 3x + 2. Multiplying by (x + 3): (x² + 3x + 2)(x + 3) = x³ + 3x² + 2x + 3x² + 9x + 6, which simplifies to x³ + 6x² + 11x + 6. Choosing x³ + 6x² + 6x + 6 has the right x² and constant terms but adds 1 + 2 + 3 = 6 for the x-coefficient instead of the correct sum of pairwise products 1×2 + 1×3 + 2×3 = 11. Choosing x³ + 5x² + 11x + 6 sums only two of the three constants (2 + 3 = 5) for the x² coefficient, leaving out the 1. Choosing x³ + 6x² + 11x + 5 adds the last two constants (2 + 3 = 5) instead of multiplying all three (1 × 2 × 3 = 6) for the constant term.
23. (a) x = 4 or x = −3 — Method: find two numbers that multiply to give −12 and add to give −1 — these are −4 and 3. So x² − x − 12 = (x − 4)(x + 3) = 0, giving x = 4 or x = −3. Distractor origins: x = −4 or x = 3 has the signs the wrong way round; x = 4 or x = 3 makes both roots positive, ignoring the sign of −12; x = 12 or x = −1 comes from reading off the coefficient and the constant directly instead of factorising.
24. (d) It crosses the x-axis at x = 3 and x = −3. — y = x² − 9 factorises as (x − 3)(x + 3), since 9 = 3², so the curve crosses the x-axis at x = 3 and x = −3. Saying it crosses once at x = 9 mistakes the constant term for a root directly, without taking its square root. Saying it crosses at x = 9 and x = −9 makes the same mistake but adds a sign either way. Saying it does not cross the x-axis confuses the y-intercept, which is negative at (0, −9), with the number of times the curve meets the x-axis — a negative y-intercept combined with an upward-opening curve guarantees it crosses the x-axis twice.