CDS Mathematics · Polynomials

The Factor Theorem

x − a is a factor of f(x) exactly when f(a) = 0; a quadratic factor gives one such equation for each of its roots.

All Polynomials notes CDS Mathematics strategy

Why this matters

Twelve PYQs. Either test which expression vanishes at a given root, or use a known factor to find unknown coefficients — one equation per root, solved together. A quadratic factor is simply two linear factors.

Concept 1 of 2: Testing for a factor

A factor

x - a

means

a

is a zero. So to test, substitute the zero and check for

0

— no division needed.

Definition

$x - a$ is a factor $\iff f(a) = 0$ ; $x + a$ is a factor $\iff f(-a) = 0$ ; $ax - b$ $\iff f\left(\dfrac ba\right) = 0$ .
$x - 1$ is a factor $\iff$ the coefficients add to $0$ .
$x + 1$ is a factor $\iff$ the even-place and odd-place coefficient sums are equal.
$x^2 - 1$ is a factor $\iff$ both of these hold.

Factor theorem

(x - a) \mid f(x) \iff f(a) = 0

Worked example

x - 3

a factor of

x^3 - 4x^2 + x + 6

? Is

x + 1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (I) 2017 — Elementary Mathematics · Q2Easy

Example 1 · Polynomials · The Factor Theorem

(x + 4)

is a factor of which one of the following expressions ?

A factor of one is not a factor of both

When asked what divides BOTH polynomials, test each candidate on each polynomial. A root of one alone is not enough.

Drill 4 more on testing for a factor

Concept 2 of 2: Finding unknown coefficients

Each known factor gives one equation. With two unknowns you need two factors — often the two linear factors of a given quadratic.

Definition

Factorise a given quadratic factor into its roots $\alpha, \beta$ ; then $f(\alpha) = 0$ and $f(\beta) = 0$ .
Solve the two linear equations for the two unknowns.
Once one zero is known, divide it out and factorise the quotient to find the rest.

Quadratic factor

(x - \alpha)(x - \beta) \mid f(x) \iff f(\alpha) = f(\beta) = 0

Worked example

x^2 - 3x + 2

is a factor of

x^4 + ax^2 + b

, find

a

and

b

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (I) 2025 — Elementary Mathematics · Q81Moderate

Example 2 · Polynomials · The Factor Theorem

x^2 - 5x + 4

is a factor of

x^4 - px^2 + q

, then what are the values of p and q respectively ?

Factorise the quadratic factor first

Substituting a quadratic like

x^2 - 4x + 3

is not possible directly; its roots

1

and

3

are what you substitute. Getting a root's sign wrong gives two consistent but wrong equations.

Drill 6 more on finding unknown coefficients

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 (2)

Testing for a factor
Factor theorem
$(x - a) \mid f(x) \iff f(a) = 0$
Finding unknown coefficients
Quadratic factor
$(x - \alpha)(x - \beta) \mid f(x) \iff f(\alpha) = f(\beta) = 0$

Watch out for (2)

A factor of one is not a factor of both
→ Testing for a factor
Factorise the quadratic factor first
→ Finding unknown coefficients

Test yourself on Polynomials

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Polynomials chapter test Drill all 12 questions