DSE Maths | Remainder and Factor Theorems: Tutor Lesson Plan

2026-11-03

Tutor lesson-prep guides · Secondary Mathematics

Students reach for long division the moment they see a cubic, then lose a sign two lines in. This lesson turns “remainder on division by x minus a” into “substitute a”, so long division becomes the backup rather than the default.

Objectives — Students can: (1) find the remainder of a cubic divided by x minus a by evaluating P(a), getting 0 for x cubed minus 4x squared plus x plus 6 at x = 3; (2) use synthetic division on 2x cubed plus x squared minus 13x plus 6 by x minus 2, obtaining the quotient 2x squared plus 5x minus 3; (3) solve two unknown coefficients from two given factors, getting a = -4 and b = 10 with the third factor x minus 5; (4) explain that the remainder theorem only applies to linear divisors, so a quadratic divisor needs two substitutions.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Substitution into a polynomial	For P(x) = x squared minus 3x plus 2, what is P(4)?	16 minus 12 plus 2, so 6
Factorising quadratics	Factorise x squared minus 5x plus 6.	(x minus 2)(x minus 3)
Dividing terms	What is 12x cubed divided by 3x?	4x squared
Remainders	What are the quotient and remainder of 17 divided by 5?	Quotient 3, remainder 2, since 17 = 5 times 3 plus 2

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opening conflict	Ask how to divide the cubic by x minus 3 without long division	Answer that substituting x = 3 is enough
8-20 min	Remainder theorem	Compare P(3) = 0 with P(half) = 45 over 8	Separate exact division from a non-zero remainder
20-32 min	Synthetic division	Lay out 2, 1, -13 and 6, divisor 2	Write the quotient 2x squared plus 5x minus 3
32-44 min	Factor theorem	From P(2) = 0 back to the factor x minus 2, then finish factorising	List all three roots: 2, a half, -3
44-54 min	Unknown coefficients	The cubic with a and b is divisible by x minus 1 and x plus 2	Get a = -4 and b = 10
54-60 min	Close	Have students check with the third root 5	Write the first exit question

03 Teaching script (read it out)

Scene 1: The remainder theorem is the definition of division, not a new formula

Write P(x) as (x minus a) times Q(x) plus R. Dividing by a linear expression leaves a constant remainder, never an x term. Substituting x = a kills the first product and leaves R. Students who memorise only “substitute a” will substitute 1 into a divisor of 2x minus 1, so make the rule explicit from the start: solve the divisor equal to zero, which gives x = a half.

Board line: the remainder is what substitution returns, not a by-product of long division.

Scene 2: Synthetic division saves paper, it changes nothing

Line up the coefficients 2, 1, -13, 6 and put 2 outside. Bring the first down, multiply by 2 to get 4 and add to 1 to get 5, multiply to get 10 and add to -13 to get -3, multiply to get -6 and add to 6 to get 0. The last cell is the remainder, the earlier cells are the quotient. Always check the degree: a cubic divided by a linear gives a quadratic.

Board line: count the degree before you compute the numbers.

Scene 3: Two factors mean two equations

If the cubic is divisible by x minus 1 and x plus 2, then P(1) = 0 and P(-2) = 0. That gives a plus b equals 6 and 4a plus b equals -6, so 3a = -12, a = -4 and b = 10. The polynomial then factorises with a third factor x minus 5, and the root 5 doubles as the check.

Board line: each factor supplies one equation, and two equations settle two unknowns.

04 Question types and worked steps

Type	Example	Steps	Answer
Find a remainder	Remainder when the cubic is divided by x minus 3	Set the divisor to zero to get x = 3, substitute, keep brackets around negatives	0, an exact division
Remainder with a coefficient	Remainder of the same cubic divided by 2x minus 1	Solve 2x minus 1 = 0 for x = a half, cube the half properly, then use a common denominator	45 over 8, that is 5.625
Factorise a cubic	Factorise 2x cubed plus x squared minus 13x plus 6	Test P(2), divide synthetically, then factorise the quadratic quotient	(x minus 2)(2x minus 1)(x plus 3)
Unknown coefficients	Find a and b if the cubic is divisible by x minus 1 and x plus 2	Substitute twice to get two equations, subtract to remove b, then verify the third root	a = -4, b = 10

Sanity check (teach this): Degree check: a cubic divided by a linear gives a quadratic quotient and a constant remainder, so a cubic quotient means a missed step. Value check: an exact division must give P(a) equal to zero, and a tiny non-zero value usually means rounding rather than divisibility. Fraction check: both the numerator and the denominator get cubed, so a half cubed is one eighth. Finally, substitute the solved coefficients back into one original equation before declaring the answer.

05 Independent practice (3 questions)

Question	Answer	What to watch
Evaluate the cubic at x = 2 and x = -3	Both are zero	Both zero means both are factors
Factorise the cubic completely	(x minus 2)(2x minus 1)(x plus 3)	Check the second bracket with x = a half
With divisibility by x minus 1 and x plus 2, find a plus b	a = -4, b = 10, so a plus b = 6	The question asks for the sum, not each value

06 Common mistakes and what to say

Mistake	Student writes	Say this
Substituting the wrong value for a scaled divisor	For 2x minus 1 I use x = 1	Set the divisor to zero: 2x minus 1 = 0 gives a half. Substituting 1 gives the remainder for x minus 1, a different division.
Leaving an x in the remainder	The remainder is 3x plus 1	A linear divisor always leaves a constant. An x term means the division is unfinished or part of the quotient was misplaced.
Forgetting to reduce the degree	The quotient is 2x cubed plus 5x minus 3	A cubic divided by a linear is quadratic, so the leading term is 2x squared. Check the degree before writing.
Using the theorem on a quadratic divisor	I substituted x = 1 into a product of two factors and got the remainder	The theorem only covers linear divisors. For a quadratic divisor the remainder can be linear, so set up two unknowns and substitute twice.

07 Homework (paper, section and question numbers)

Maths chapter index — Revise quadratic equations and the discriminant, then do questions 1 and 2
Quadratic equations and the discriminant — Do question 3 on finding the other root, then questions 4 to 6
Graph transformations — Optional: link the graph of P(x) to why P(a) = 0 means an intercept

08 Exit check (last 5 minutes)

Question	Answer	Decision
Remainder of the cubic divided by x minus 3	0	Error: substituted -3
Remainder of the same cubic divided by 2x minus 1	45 over 8	Error: substituted 1
Full factorisation of the cubic	(x minus 2)(2x minus 1)(x plus 3)	Error: stopped at x minus 2
Degree of a remainder for a quadratic divisor	At most one, so ax plus b	Error: said always constant

09 Tutor FAQ

Why does substituting a give the remainder?

Because P(x) equals (x minus a)Q(x) plus R, and the first term vanishes at x = a. It is the identity itself, not a trick.

When should I use long division instead?

Use synthetic division for linear divisors, because it only works there. For quadratic or higher divisors, long division or comparison of coefficients is the reliable route.

Do unknown coefficients always need two equations?

Two unknowns need two conditions, so the question always supplies two: two factors, two remainders, or one of each.

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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.