CDS Mathematics · Polynomials

The Remainder Theorem

The remainder when f(x) is divided by x − a is f(a); on division by a quadratic the remainder is linear and is fixed by its values at the two roots.

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Why this matters

Fifteen PYQs, most EASY or MODERATE. Substituting one number replaces long division every time. The harder items divide by a quadratic — find the linear remainder from two values — or use the fact that xⁿ − aⁿ always has the factor x − a.

Concept 1 of 3: Remainder on division by a linear factor

Write

f(x) = (x - a)q(x) + r

. Putting

x = a

kills the first term, so the remainder is just

f(a)

. For

ax - b

the root is

\dfrac ba

Definition

Dividing by $x - a$ : remainder $f(a)$ . Dividing by $x + a$ : remainder $f(-a)$ . Dividing by $ax - b$ : remainder $f\left(\dfrac ba\right)$ .
A given remainder is one equation for an unknown coefficient.
Equal remainders for $x - 1$ and $x + 1$ : $f(1) = f(-1)$ , which kills every odd-power coefficient.
The quotient comes from synthetic division, writing $0$ for any missing power.

Remainder theorem

f(x) \div (x - a) \;\Rightarrow\; \text{remainder} = f(a)

Worked example

When

2x^3 - x^2 + kx + 4

is divided by

x - 2

the remainder is

18

. Find

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (I) 2018 — Elementary Mathematics · Q16Easy

Example 1 · Polynomials · The Remainder Theorem

The remainder when

3x^3 + kx^2 + 5x - 6

is divided by

(x + 1)

-7

. What is the value of

k

x + a means substitute −a

Dividing by

x + 3

leaves

f(-3)

, not

f(3)

. The wrong sign gives a remainder of the right size and wrong value, and it is usually printed.

Drill 6 more on remainder on division by a linear factor

Concept 2 of 3: Remainder on division by a quadratic

Dividing by a quadratic leaves a remainder of degree at most one,

r(x) = px + q

. The remainder agrees with

f

at the two roots of the divisor, which gives two equations for

p

and

q

Definition

$f(x) = (x - \alpha)(x - \beta)Q(x) + (px + q)$ , so $p\alpha + q = f(\alpha)$ and $p\beta + q = f(\beta)$ .
Dividing by $x^2 + 1$ : replace every $x^2$ by $-1$ .
A polynomial divisible by $x^2 + 1$ and $x^4 + 1$ has both as factors; look for the factorisation first.

Linear remainder

r(x) = \dfrac{(x - \beta)f(\alpha) - (x - \alpha)f(\beta)}{\alpha - \beta}

Worked example

f(x)

leaves remainder

5

on division by

x - 2

and

-1

on division by

x + 1

. Find the remainder on division by

(x - 2)(x + 1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (II) 2025 — Elementary Mathematics · Q1Moderate

Example 2 · Polynomials · The Remainder Theorem

Let

p(x)

be a polynomial. When

p(x)

is divided by

(x-1)

, it leaves 2 as the remainder. When

p(x)

is divided by

(x-2)

, it leaves 1 as the remainder. What is the remainder when

p(x)

is divided by

(x-1)(x-2)

The remainder is not a number

On division by a quadratic the remainder is usually a linear expression like

3 - x

. An option that is a plain number fits only if the two values happen to be equal.

Drill 4 more on remainder on division by a quadratic

Concept 3 of 3: Divisibility of xⁿ ± aⁿ

Put

x = a

into

x^n - a^n

and you get zero for every

n

, so

x - a

always divides it. Put

x = -a

into

x^n + a^n

: the result is zero only when

n

is odd.

Definition

$x - a$ divides $x^n - a^n$ for every natural $n$ .
$x + a$ divides $x^n - a^n$ when $n$ is even.
$x + a$ divides $x^n + a^n$ when $n$ is odd, never when $n$ is even.
$x^{2n} - y^{2n} = (x^n - y^n)(x^n + y^n)$ , so any expression 'difference of even powers plus a constant' leaves that constant as remainder.

Factor of xⁿ − aⁿ

x^n - a^n = (x - a)\left(x^{n - 1} + x^{n - 2}a + \cdots + a^{n - 1}\right)

Worked example

Which of

x^6 - 1

x^6 + 1

x^5 + 1

are divisible by

x + 1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q10Easy

Example 3 · Polynomials · The Remainder Theorem

(x^n - a^n)

is divisible by

(x - a)

, where

x \ne a

, for every

Even n and a plus sign

x^n + y^n

with

n

even is never divisible by

x + y

: at

x = -y

it equals

2y^n

. Knowing

n

is even therefore answers the question — with a 'no'.

Drill 2 more on divisibility of xⁿ ± aⁿ

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Remainder on division by a linear factor
Remainder theorem
$f(x) \div (x - a) \;\Rightarrow\; \text{remainder} = f(a)$
Remainder on division by a quadratic
Linear remainder
$r(x) = \dfrac{(x - \beta)f(\alpha) - (x - \alpha)f(\beta)}{\alpha - \beta}$
Divisibility of xⁿ ± aⁿ
Factor of xⁿ − aⁿ
$x^n - a^n = (x - a)\left(x^{n - 1} + x^{n - 2}a + \cdots + a^{n - 1}\right)$

Watch out for (3)

x + a means substitute −a
→ Remainder on division by a linear factor
The remainder is not a number
→ Remainder on division by a quadratic
Even n and a plus sign
→ Divisibility of xⁿ ± aⁿ

Test yourself on Polynomials

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