中文（繁體）
中文（简体）
English
DSE Maths | Polynomial Division and the Remainder Theorem — Teacher Prep Guide | Three moves, remainder 1 and three roots
Students skip any cubic divided by a linear term, then lose the first mark by writing a quotient of the wrong degree. This lesson reduces division to three hand moves and two theorems.
Objectives — Students will be able to: (1) divide a degree-3 polynomial by a linear divisor and state a degree-2 quotient with its remainder; (2) use the remainder theorem to get the remainder by substitution and explain why no division is needed; (3) use the factor theorem to test a linear factor and factorise a cubic completely into three linear factors.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Degree | What is the degree of 3x^3 – 5x^2 + 2x – 7 | 3, the highest power |
| Substitution | Evaluate 3x^3 at x = 2 | 24 |
| Basic factorising | Factorise x^2 – 2x – 3 | (x – 3)(x + 1) |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Two methods | Do the same question by long division and by substitution | Compare the remainders |
| 8-20 min | Three moves | Divide, multiply, subtract, bring down | Divide it themselves |
| 20-32 min | Remainder theorem | Dividing by x – a leaves p(a); check p(2) = 1 | Three substitutions aloud |
| 32-44 min | Divisor ax – b | For 2x – 1 substitute 0.5, remainder 4 | One on their own |
| 44-56 min | Factor theorem | Test plus and minus 1, 2, 3 in x^3 – 4x^2 + x + 6 | Pairs share the tests |
| 56-60 min | Exit check | Collect and mark three questions | Work alone |
03 Teaching script (read it out)
Scene 1: Long division is three moves: divide, multiply, subtract
Divide: 3x^3 by x gives 3x^2, written in the quotient. Multiply: 3x^2 times (x – 2) gives 3x^3 – 6x^2. Subtract: (3x^3 – 5x^2) minus (3x^3 – 6x^2) leaves x^2, then bring down +2x.
Three rounds give quotient 3x^2 + x + 4 and remainder 1. Missing terms must be written as zero (x^3 + 1 becomes x^3 + 0x^2 + 0x + 1), otherwise every column is misaligned.
Three rounds give quotient 3x^2 + x + 4 and remainder 1. Missing terms must be written as zero (x^3 + 1 becomes x^3 + 0x^2 + 0x + 1), otherwise every column is misaligned.
Ask students: write each multiplication on its own line, aligned by power; misalignment means starting again.
Scene 2: The remainder theorem turns division into one substitution
Dividing p(x) by (x – a) leaves p(a). Check with the same question: p(2) = 3(8) – 5(4) + 2(2) – 7 = 1, matching the division.
One step further: for a divisor ax – b, solve ax – b = 0 to get x = b/a, so the remainder is p(b/a). For 2x – 1 substitute 0.5. On 2x^3 + 3x^2 – 4x + 5, p(0.5) = 4; not -0.5 and not 1.
One step further: for a divisor ax – b, solve ax – b = 0 to get x = b/a, so the remainder is p(b/a). For 2x – 1 substitute 0.5. On 2x^3 + 3x^2 – 4x + 5, p(0.5) = 4; not -0.5 and not 1.
Ask students: why the lower the divisor degree the more time substitution saves, and what shape the remainder would take for a quadratic divisor.
Scene 3: The factor theorem: a zero remainder means a factor
From the remainder theorem: p(a) = 0 exactly when (x – a) is a factor. Test x^3 – 4x^2 + x + 6: p(2) = 8 – 16 + 2 + 6 = 0, so (x – 2) is a factor and division gives x^2 – 2x – 3, which factorises to (x – 3)(x + 1). The roots are 2, 3 and -1.
Check: the product of the roots is -6, which equals minus the constant term divided by the leading coefficient. That check takes thirty seconds and catches any wrong root.
Check: the product of the roots is -6, which equals minus the constant term divided by the leading coefficient. That check takes thirty seconds and catches any wrong root.
Ask students: list candidates from the factors of 6 (plus and minus 1, 2, 3, 6) before substituting anything.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Long division | Cubic divided by a linear term | Divide, multiply, subtract, bring down | Quotient degree is one less |
| Remainder theorem | Remainder for x – a or ax – b | Solve the divisor for a, then substitute | For 2x – 1 use 0.5 |
| Factor theorem | Prove a factor and factorise fully | Zero gives a factor, divide, then split again | List every root at the end |
| Unknown coefficient | Find k from a given remainder | Write p(a) in terms of k, set it equal | Remainders may be negative |
Sanity check (teach this): Sanity checks: the quotient is quadratic because a cubic is divided by a linear term; the remainder must be a constant and cannot contain x; p(2) = 1 agrees with the division; the roots 2, 3 and -1 multiply to -6, matching the constant term of 6.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Long division | Divide 2x^3 + x^2 – 5x + 3 by x + 1 | Align every column, fill missing terms |
| Remainder theorem | Same question by substitution | Both answers must agree |
| Factor theorem | Factorise x^3 – 2x^2 – 5x + 6 | Candidates plus and minus 1, 2, 3, 6 |
| Unknown coefficient | p(x) = x^3 + kx – 4 leaves 6 when divided by x – 2 | Write p(2) and solve |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Misaligned division | Putting the 3x^2 product in the x^2 column | x cubed under x cubed, x squared under x squared. One bad row ruins the rest. |
| Wrong value for ax – b | Substituting -0.5 into 2x – 1 | Solve 2x – 1 = 0 first. What is x? Substitute that. |
| A remainder with x in it | Answering the remainder as x + 1 | A linear divisor leaves a constant. If x survives, you have not divided enough. |
| Stopping at one root | Writing (x – 2) and claiming done | Fully factorised means three linear factors. The quotient is still quadratic. |
07 Homework (paper, section and question numbers)
- Secondary Maths chapter index — Every chapter in the subject
- All prep guides — Parallel chapters in other subjects
- Redo this chapter — Two divisions and two factorisations, showing every line
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Remainder when dividing by x – 2 | p(2), here 1 | Answering divide first means restart act one |
| Value to substitute for 2x – 1 | x = 0.5 | Answering -0.5 means redo act two |
| Roots of x^3 – 4x^2 + x + 6 | 2, 3 and -1 | One root only means redo act three |
09 Tutor FAQ
Must students write long division?
The theorem is faster but gives no quotient. To factorise you must divide, so the two work together.
Can they just use the calculator to find roots?
Fine for checking, but the marks are in listing candidates and substituting. Always start from the factors of the constant term.
Does every cubic have three real roots?
No. There may be a repeated root or one real and two complex roots; DSE questions usually design three real roots, so answer to the question asked.
Find a tutor — freeTutor Registration →
A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.