F6 MATHEMATICS · DSE REVISION

Polynomials

In every DSE since 2012. About eight marks. Today we make them safe.

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PAPER 1 MARKS ON POLYNOMIALS, RECENT PAPERS

0Paper 1 marks in the last ten papers, plus one MC in Paper 2 every year

Never skipped. Always the same shapes.

Every year of the DSE

Level 5 means not one mark dropped here.

3

tools

division algorithm, remainder and factor theorems

4

MC shapes

every recent Paper 2 question

4

ways in

Paper 1 part (a)

1

method

for every question on the nature of the roots

2

twists

the 2023 and 2026 Level 5 parts

F6 Mathematics · Polynomials

Today

Division algorithm
\f(x) = \text{divisor}\times\text{quotient} + \text{remainder}
Remainder theorem
\f(x) \div (ax - b) leaves \f\left(\tfrac{b}{a}\right)
Factor theorem
ax - b is a factor \iff \f\left(\tfrac{b}{a}\right) = 0

Degree rule. Divide by a quadratic: the remainder is mx + n. Cubic ÷ quadratic: the quotient is mx + n. Quadratic ÷ quadratic: the quotient is a constant k.

F6 Mathematics · Polynomials

1 | Three tools

2026 P1 Q14(a)When \g(x) is divided by x^2 + 2x + a, the quotient and remainder are x + 13 and b(x - 3). \g(x) is divisible by x - 3. Find a.

\g(x) = (x^2 + 2x + a)(x + 13) + b(x - 3)

Which value of x makes one term vanish?

\g(3) = (9 + 6 + a)(16) + b(3 - 3) = 0
16(15 + a) = 0 \implies a = -15  two unknowns, one equation, still solved
F6 Mathematics · Polynomials

1 | Substitute the root

Paper 2: one MC, four shapes.

A

Numbers

Factor gives c, then substitute. 2017, 2020, 2025

B

Unknown in divisor

x + k: substitute x = -k. 2019, 2022, 2024

C

Too many unknowns

Do not solve. Odd powers flip. 2018, 2023, 2026

D

Quadratic divisor

Remainder ax + b, two roots. 2021

F6 Mathematics · Polynomials

2 | Paper 2

2026 P2 Q8

Let \g(x) = hx^3 + 2x^2 + kx - 5, where h and k are constants. If \g(x) is divisible by x - 4, find the remainder when \g(x) is divided by x + 4.

Two unknowns, one equation. The next screen shows the answer. Has everyone chosen?

F6 Mathematics · Polynomials

2 | Shape C

Do not solve. Flip.

hx^32x^2kx-5
x = 464h+32+4k-5= 0
x = -4-64h+32-4k-5= \;?
From the first row: 64h + 4k = -27
\g(-4) = -(64h + 4k) + 27 = 27 + 27 = 54  D

Odd powers change sign, even powers do not. The same move works in 2018 and 2023.

F6 Mathematics · Polynomials

2 | Shape C

Alone

Worksheet A1 to A4: one question of each shape. About 75 seconds each, as in the exam.

Check

Before you choose: what value of x did you substitute? 3x - 1 means x = \tfrac13.

The next screen shows the answers. Has everyone circled four letters?

F6 Mathematics · Polynomials

2 | Your turn

QAnswerThe line that gets it
A1B\p(-3) = 0 \Rightarrow c = -12; \p\left(\tfrac13\right) = -10
A2C\f(-k) = 2k + 8 = 0 \Rightarrow k = -4; \f(-1) = 5
A3D32a + 2b = -9; \f(-2) = -(32a + 2b) + 9 = 18
A4Bremainder mx + n: 2m + n = 9,\; -m + n = 0 \Rightarrow 3x + 3

Wrong options are sign errors. Check the substituted value, every time.

F6 Mathematics · Polynomials

2 | Answers

Paper 1 (a): which way in?

  1. 2025\f(x) = 2x^3 + hx^2 + kx + 15. Remainder -45 when divided by x + 2; divisible by 2x - 5.
  2. 2021\p(x) is divisible by x - 5. Divided by x^2 + x + 1: quotient 2x^2 - 37, remainder cx + c - 1.
  3. 2022\p(x) = 2x^3 + ax^2 + bx - 20. Divided by x^2 - 2x + 3, the remainder is x + 13.
  4. 2019Cubic \p(x) divisible by 2x^2 + 9x + 14; remainders 50 and -52 for x - 1 and x + 2.

Write only the first line for each. The next screen shows them.

F6 Mathematics · Polynomials

3 | Four ways in

Way inFirst line (the 1M)
1Remainder + factor theorems\f(-2) = -45,\; \f\left(\tfrac52\right) = 0: two equations
2Division algorithm, substitute the root\p(5) = (31)(13) + 6c - 1 = 0: the quotient part is just a number
3Compare coefficients\p(x) = (mx + n)(x^2 - 2x + 3) + x + 13: m = 2, 3n + 13 = -20
4Unknown quotient\p(x) = (ax + b)(2x^2 + 9x + 14); substitute 1 and -2

Read what is given. The given information chooses the method.

F6 Mathematics · Polynomials

3 | Four ways in

The last part asks the same thing every year: what kind of numbers are the roots?

Real
Rational
Integer 3 0
\tfrac{5}{2} -\tfrac43
\tfrac{-5 \pm \sqrt{73}}{2}
Not real roots of 2x^2 + 9x + 14 = 0
\Delta = -31 \lt 0

Irrational means real but not rational. Roots that are not real are not irrational.

F6 Mathematics · Polynomials

4 | The nature of the roots

\f(x) = (x - 3)(4x^2 + 20x + 45), and for the quadratic \Delta = 400 - 720 = -320.

How many irrational roots does f(x) = 0 have?

This exact trap was set in 2018 and again in 2022. The next screen shows the answer. Has everyone chosen?

F6 Mathematics · Polynomials

4 | Trap

0

The quadratic has no real roots, so no irrational roots. And 3 is not irrational.

Say both halves in your answer. That is where the last mark is.

4 | Trap

2022 P1 Q14(c)Someone claims that \p(x) = 2x^3 - 15x^2 + 29x - 20 = 0 has two irrational roots. Do you agree?

Factorise  (x - 5)(2x^2 - 5x + 4) = 01M Settle the quadratic  \Delta = (-5)^2 - 4(2)(4) = -7 \lt 01M Check every root  The quadratic has no real roots, hence no irrational roots. Note that 5 is not irrational. Conclude  Thus, the claim is disagreed.1A f.t.

Same three marks every year. A wrong (a) can still score them by follow-through, so always finish.

F6 Mathematics · Polynomials

4 | The method

Alone

Worksheet B2, exam conditions. 8 marks, 7 minutes. No talking, no notes.

Swap

Swap with your neighbour. Mark their work with the scheme on the next screen: M for the method, A for the answer.

The next screen shows the marking scheme. Pens down?

F6 Mathematics · Polynomials

5 | Exam conditions

(a) \p(x) = (x^2 + ax - 6)(2x + 1) + k(x + 3)1M \p(-3) = 01M (9 - 3a - 6)(-5) = 0 \Rightarrow a = 11A (b) \p(x) = (x + 3)(x - 2)(2x + 1) + k(x + 3)
\phantom{\p(x)} = (x + 3)(2x^2 - 3x - 2 + k)1M \p(1) = 4(k - 3) = -16 \Rightarrow k = -11A
(b) continued x = -3 or 2x^2 - 3x - 3 = 01M x = \dfrac{3 \pm \sqrt{33}}{4}1M \tfrac{3 \pm \sqrt{33}}{4} are irrational; -3 is not. Thus, \p(x) = 0 has 2 irrational roots.1A f.t. Lost the last mark? Check the conclusion sentence first.
F6 Mathematics · Polynomials

5 | Marking scheme

2026 P1 Q14(b)

From (a): \g(x) = (x - 3)(x^2 + 18x + 65 + b)

Is there only one value of b such that at least two roots of g(x) = 0 are equal?

The next screen shows the answer. Has everyone written a reason, not only a guess?

F6 Mathematics · Polynomials

6 | Level 5 twist

Two ways for roots to be equal.

Case 1
The quadratic has equal roots: \Delta = 18^2 - 4(65 + b) = 0 \Rightarrow b = 16
Case 2
The quadratic also has the root 3: 9 + 54 + 65 + b = 0 \Rightarrow b = -128

No. b = 16 or b = -128. The trap is stopping after Case 1. Worksheet B5 is the same shape.

F6 Mathematics · Polynomials

6 | Level 5 twist

2023 P1 Q13When \h(x) is divided by \g(x), the quotient and the remainder are equal. Call them both Q.

\h(x) = Q\,\g(x) + Q
\phantom{\h(x)} = Q\,\big(\g(x) + 1\big)  factor it out
2023: \g(x) + 1 = x^3 + 5x^2 - 12x = x(x^2 + 5x - 12), so 0 is a root. Count it: 0 is rational.

Worksheet B4 is the same shape.

F6 Mathematics · Polynomials

6 | Level 5 twist

Six ways to lose a mark you earned.

1

\tfrac12 is rational but not an integer. Read the claim word by word. (2017)

2

\Delta \lt 0: no real roots, so no irrational roots either. (2018, 2022)

3

0 is a rational root. Count it. (2023)

4

Check the linear factor's root too. It can be the repeated one. (2026)

5

No conclusion sentence, no 1A. "Thus, the claim is disagreed."

6

"It cannot be factorised" is not a reason. Show \Delta or the formula.

F6 Mathematics · Polynomials

7 | Traps

Exit ticket

Let \f(x) = ax^3 + 2x^2 + bx + 1, where a and b are constants. If \f(x) is divisible by x - 3, find the remainder when \f(x) is divided by x + 3.

On a slip of paper: your letter and the one line that got it. The next screen shows the answer.

F6 Mathematics · Polynomials

8 | Exit ticket

D27a + 3b = -19, so \f(-3) = -(27a + 3b) + 18 + 1 = 38

Homework: the rest of the worksheet.

A5 to A8 at 75 seconds each. B1, B3 to B6 at one minute per mark, on foolscap. Revise from the notes and the drill on the reading page.

Polynomials: done