1 · Recent papers at a glance
| Year | P1 | Marks | Part (a): way in | Last part asks | P2 shape |
|---|---|---|---|---|---|
| 2017 | Q14 | 8 | Compare coefficients | roots of \f - \g = 0 all integers? | A |
| 2018 | Q12 | 7 | Remainder + factor theorems | at least one irrational root? | C |
| 2019 | Q11 | 6 | Unknown quotient ax + b | how many rational roots? | B |
| 2020 | Q13 | 8 | Division algorithm, substitute a root | all roots integers? | A |
| 2021 | Q12 | 7 | Division algorithm, substitute a root | prove a factor; all roots real? | D |
| 2022 | Q14 | 8 | Compare coefficients | is it a factor? two irrational roots? | B |
| 2023 | Q13 | 7 | Compare coefficients (quotient = remainder) | how many rational roots? | C |
| 2024 | Q14 | 8 | Constant term, then remainder + factor theorems | how many irrational roots? | B |
| 2025 | Q10 | 6 | Remainder + factor theorems | all roots rational? | A |
| 2026 | Q14 | 8 | Division algorithm, substitute a root | equal roots: only one value of b? | C |
The pattern: part (a) finds the unknown constants (3 marks). In every paper above, the last part asks about the nature of the roots: integer, rational, irrational, real, or equal.
2 · The three tools
Division algorithm
\f(x) = d(x)\,Q(x) + R(x), where \deg R \lt \deg d.
Remainder theorem
\f(x) \div (ax - b) leaves the remainder \f\left(\tfrac{b}{a}\right). For 2x + 1, substitute x = -\tfrac12.
Factor theorem
ax - b is a factor of \f(x) if and only if \f\left(\tfrac{b}{a}\right) = 0. "Divisible by" means "is a factor of".
The degree rule. Divide by a quadratic and the remainder is mx + n. A cubic divided by a quadratic has a linear quotient mx + n. A quadratic divided by a quadratic has a constant quotient k.
3 · Paper 2: one MC, four shapes
Every recent MC has the same idea: a factor gives you an equation, then you find a remainder or a value.
| Shape | Real example | Key move |
|---|---|---|
| A Numbers | \p(x) = 2x^2 - 11x + c divisible by x - 7; remainder when divided by 2x + 1? 2017 · also 2020, 2025 | \p(7) = 0 \Rightarrow c = -21, then \p\left(-\tfrac12\right) = -15. |
| B Unknown in the divisor | x^3 + kx^2 + 5x + 10 divisible by x + k 2024 · also 2019, 2022 | Substitute x = -k. The k^3 terms cancel: k = 2. |
| C Too many unknowns | \g(x) = hx^3 + 2x^2 + kx - 5 divisible by x - 4; remainder when divided by x + 4? 2026 · also 2018, 2023 | Do not solve. 64h + 4k = -27. Odd powers flip sign, even powers do not: \g(-4) = -(64h + 4k) + 27 = 54. |
| D Quadratic divisor | \p(-1) = -2, \p(1) = 0; remainder when divided by x^2 - 1? 2021 | Let the remainder be ax + b; substitute both roots: x - 1. |
4 · Paper 1 part (a): four ways in
Read what the question gives you, then write the first line that earns the method mark.
| The question gives | Your first line | Years |
|---|---|---|
| Remainders and a factor, with \f(x) written out | \f(-2) = -45 and \f\left(\tfrac52\right) = 0: simultaneous equations | 2018, 2024, 2025 |
| "divided by …, quotient …, remainder …" and a factor x - r | \f(x) = d(x)Q(x) + R(x), then substitute x = r: one term vanishes or becomes a plain number | 2020, 2021, 2026 |
| Divisor and remainder; \f(x) known except a few coefficients | \f(x) = (mx + n)\,d(x) + R(x); compare the highest and the constant terms | 2017, 2022, 2023 |
| Divisible by a quadratic, remainders for two linear divisors | \p(x) = (ax + b)\,d(x); substitute | 2019 |
5 · The last part: the nature of the roots
- Factorise into (linear) × (quadratic), using part (a) and the known factor. 1M
- Settle the quadratic with \Delta = b^2 - 4ac or the quadratic formula. 1M
- Check every root, the linear factor's too, against the property asked; write the conclusion sentence. 1A f.t.
The number ladder
| integers | \dots, -2, 0, 3, \dots |
| ⊂ rationals | also \tfrac23,\ -\tfrac43 |
| ⊂ reals | also irrationals: \tfrac{-5 \pm \sqrt{73}}{2} |
| not real | roots when \Delta \lt 0 |
What \Delta tells you
| \Delta \lt 0 | no real roots, so none rational or irrational |
| \Delta = 0 | two equal rational roots |
| \Delta \gt 0, perfect square | two distinct rational roots |
| \Delta \gt 0, not a square | two irrational roots |
For integer coefficients, as in every DSE question.
Conclusion sentences that score
Claim. "Note that \tfrac12 is a root of the equation, and it is not an integer. So not all the roots are integers. Thus, the claim is disagreed."
How many. "Note that \tfrac{-1 \pm \sqrt{97}}{12} are irrational, and 2 and -5 are not irrational. Thus, the equation has 2 irrational roots."
Is it a factor. "\p(5) = 0. Thus, x - 5 is a factor of \p(x)."
6 · The Level 5 twists
| Twist | The move |
|---|---|
| Quotient = remainder 2023 | \h = Q\,\g + Q = Q(\g + 1). Factorise \g + 1; in 2023 it had a factor x, so 0 was a root. |
| Equal roots 2026 | \g(x) = (x - 3)(x^2 + 18x + 65 + b). Two cases: \Delta = 0 gives b = 16; 3 also a root of the quadratic gives b = -128. So not only one value. |
| An unknown \g(x) 2017 | \g is quadratic with the same remainder, so \g = k\,d(x) + R(x) and \f - \g = d(x)(Q(x) - k). One bad root disproves "all"; you never need k. |
| A remainder used twice 2020 | The remainder on division by x^2 - 1 also gives \f(-1) free: substitute x = -1. |
| Quartic as a product 2024 | Constant term of (6x^2 + x + p)(qx^2 + rx - 10) is -10p. Factorise the quadratic that splits; use the formula on the other. |
7 · Traps that cost marks
- \tfrac12 is rational but not an integer (2017). Read the claim word by word.
- \Delta \lt 0 does not mean irrational. No real roots, so no irrational roots either (2018, 2022).
- 0 is a rational root (2023). Count it.
- Check the linear factor's root too. In an equal-roots question it can be the repeated one (2026).
- No conclusion sentence, no 1A. "Thus, the claim is disagreed."
- "It cannot be factorised" is not a reason. Show \Delta or the formula; that line is the method mark.
8 · MC drill
Endless Paper 2 twins, new numbers every time. Aim for 75 seconds each, and a streak of 10.
9 · In the exam
- Time. About 75 seconds for the MC; about one minute per mark in Paper 1.
- Check with the calculator. Once you have the constants, use CALC to evaluate \f(r) at each root you found. Each should give 0.
- Follow-through. A wrong constant in part (a) can still earn the method marks and the 1A f.t. later. Always finish the question.