The Factor and Remainder Theorems (Pure Mathematics 3 – 9709) — Mathematics

Introduction

The Factor and Remainder Theorems are powerful algebraic tools that allow you to analyse polynomials without performing full long division every time. In the 9709 exam, they appear regularly in questions asking you to:

find an unknown coefficient in a polynomial given a known factor or remainder,
factorise a cubic or quartic fully and hence solve a polynomial equation,
calculate the remainder when a polynomial is divided by a linear expression of the form $(ax + b)$ .

The syllabus explicitly requires fluency with factors of the form $(ax + b)$ where $a \neq 1$ , so particular attention is given to that case throughout this note.

Core Concept

The Remainder Theorem

When a polynomial $f(x)$ is divided by a linear divisor $(ax + b)$ , the remainder is the constant value obtained by substituting $x = -\dfrac{b}{a}$ into $f(x)$ .

This follows from the division algorithm: if $f(x)$ is divided by $(ax+b)$ , then

f(x) = (ax + b)\,q(x) + R

where $q(x)$ is the quotient polynomial and $R$ is a constant remainder. Setting $ax + b = 0$ , i.e. $x = -\dfrac{b}{a}$ , makes the $(ax+b)$ term vanish, giving directly $R = f\!\left(-\dfrac{b}{a}\right)$ .

The Factor Theorem

The Factor Theorem is a special case of the Remainder Theorem. $(ax + b)$ is a factor of $f(x)$ if and only if the remainder is zero, i.e.

f\!\left(-\frac{b}{a}\right) = 0

This is an if and only if statement: finding that $f\!\left(-\frac{b}{a}\right) = 0$ guarantees $(ax+b)$ divides $f(x)$ exactly, and vice versa.

Why the substitution is $x = -b/a$

Setting $ax + b = 0$ solves to $x = -\dfrac{b}{a}$ . For example:

$(x - 3)$ : substitute $x = 3$
$(2x + 1)$ : substitute $x = -\tfrac{1}{2}$
$(3x - 2)$ : substitute $x = \tfrac{2}{3}$

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