Learn7357Paper 1 · PureAlgebra and functionsPolynomials: division, factor and remainder theorems

Algebra and functions · Subtopic 6 of 9

Polynomials: Division, Factor and Remainder Theorems

Long division for polynomials, why the remainder on dividing by $(x-a)$ is just $p(a)$ , and how the factor theorem lets you factorise cubics and find unknown coefficients.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 45 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Divide a polynomial by a linear or quadratic polynomial to find the quotient and remainder.
Use the remainder theorem: the remainder on dividing $p(x)$ by $(x-a)$ is $p(a)$ , and by $(ax-b)$ is $p\left(\tfrac ba\right)$ .
Use the factor theorem to test for and find factors.
Form and solve equations for unknown coefficients, then factorise fully and solve.

Pause and think

$17\div5=3$ remainder $2$ , so $17=5\times3+2$ . Write the same kind of statement for dividing $p(x)$ by $(x-2)$ .

Section 2 of 14

Core idea

Start hereTesting a guess without dividing

Does $(x-3)$ divide exactly into $2x^3-3x^2-11x+6$ ? You could do long division. Or you could just put $x=3$ in: $54-27-33+6=0$ . Zero means yes. This shortcut, the factor theorem, turns cubic equations from impossible into routine.

The big idea

Remainder theorem: dividing $p(x)$ by $(x-a)$ leaves remainder $p(a)$ . Factor theorem: $(x-a)$ is a factor ⇔ $p(a)=0$ . Use them to factorise and solve polynomials.

1. Polynomial long division

p(x)=(\text{divisor})\times q(x)+r

$p(x)$: the polynomial being divided (dividend).
$q(x)$: the quotient.
$r$: the remainder (degree less than the divisor).

Worked example $(2x^3-3x^2-11x+6)\div(x-3)$

Divide.

Goal: Quotient and remainder.
Strategy: Divide leading terms, multiply back, subtract, repeat.

$\tfrac{2x^3}x=2x^2;\ 2x^2(x-3)=2x^3-6x^2;\ \text{subtract: }3x^2-11x$
Step 1.
$\tfrac{3x^2}x=3x;\ 3x(x-3)=3x^2-9x;\ \text{subtract: }-2x+6$
Step 2.
$\tfrac{-2x}x=-2;\ -2(x-3)=-2x+6;\ \text{subtract: }0$
Step 3.

AnswerQuotient

2x^2+3x-2

, remainder

0

; so

p(x)=(x-3)(2x-1)(x+2)

✓ Check

(x-3)(2x^2+3x-2)=2x^3+3x^2-2x-6x^2-9x+6=2x^3-3x^2-11x+6

✓.

2. The remainder and factor theorems

Proof of the remainder theorem

$p(x)=(x-a)q(x)+r$
Division by a linear term leaves a constant remainder.
$p(a)=(a-a)q(a)+r$
Substitute $x=a$ (true for all $x$ , so for $a$ ).
$p(a)=0+r=r$
$a-a=0$ .

Remainder

=p(a)

. If

p(a)=0

, the remainder is

0

, so

(x-a)

is a factor (factor theorem). Dividing by

(ax-b)

: remainder

p\left(\frac ba\right)

Factors and roots: y = (x − 1)(x − 3)(x + 2)

Each factor

(x-a)

gives a crossing at

x=a

. A cubic with a positive

x^3

term starts low on the left and ends high on the right.

Worked exampleSolve $x^3-2x^2-5x+6=0$

Solve.

Goal: All roots.
Strategy: Find one root by testing factors of $6$ ; divide; solve the quadratic.

$p(1)=1-2-5+6=0$
Try $\pm1,\pm2,\pm3,\pm6$ .
$p(x)=(x-1)(x^2-x-6)$
Divide (or compare coefficients).
$=(x-1)(x-3)(x+2)$
Factorise the quadratic.

Answer

x=1,3,-2

✓ Check

p(3)=27-18-15+6=0

✓.

Worked exampleUnknown coefficients

$p(x)=x^3+ax^2+bx-6$ ; $(x-2)$ is a factor and $p(x)\div(x+1)$ leaves remainder $-12$ . Find $a$ , $b$ .

Goal: $a$ , $b$ .
Strategy: Two facts ⇒ two equations.

$p(2)=8+4a+2b-6=0\Rightarrow2a+b=-1$
Factor theorem.
$p(-1)=-1+a-b-6=-12\Rightarrow a-b=-5$
Remainder theorem.
$3a=-6\Rightarrow a=-2,\ b=3$
Add the equations.

Answer

a=-2

b=3

✓ Check

p(2)=8-8+6-6=0

✓.

Common misconceptionSign of the root For the factor

(x+3)

test

p(-3)

, not

p(3)

Pause and think

Show $(x+3)$ is a factor of $x^3+x^2-4x+6$ .

Summary

Long division: divide, multiply, subtract.
Remainder $=p(a)$ ; factor ⇔ $p(a)=0$ .

Factor theorem: $(x-a)$ is a factor iff $p(a)=0$ .
Quotient: Result of division.

HistorySolving cubics by formula. In 1545 Gerolamo Cardano published a formula for solving any cubic equation, using a method he had obtained from Niccolò Tartaglia after promising to keep it secret. Tartaglia was furious. The formula is so complicated that factorising with the factor theorem is still the method of choice whenever there is a nice root.

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