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Polynomials and the Factor Theorem

Q: What is the factor theorem in simple terms?

If substituting x = p into a polynomial gives zero, then (x - p) divides it exactly — it is a factor. It lets you find a factor by testing values rather than guessing brackets.

Q: How do I know which values to test?

Try the factors of the constant term first: for x^3 - 6x^2 + 11x - 6, test 1, 2, 3, 6. A whole-number root, if there is one, is almost always among them.

Q: What is the difference between the factor and remainder theorems?

The remainder theorem says f(p) is the remainder when you divide by (x - p). The factor theorem is the special case where that remainder is zero — meaning (x - p) is a factor.

The factor theorem is a shortcut for factorising cubics and higher polynomials: if substituting $x = p$ makes the polynomial zero, then $(x - p)$ is a factor. It turns a search into a quick test.

30 min Video by TLMaths Algebra and Functions

The big picture

Quadratics factorise into two brackets, but how do you factorise a cubic like $x^{3} - 6 x^{2} + 11 x - 6$ ? Trial factorising is hopeless. The factor theorem gives you a way in: because a root and a factor are the same information seen from two angles, you can $test$ small values until one makes the polynomial zero, and that instantly hands you a factor. Combined with algebraic division to strip that factor out, you can break any nice polynomial down to its roots. This underlies solving cubic equations, sketching polynomial curves, and later work in partial fractions and integration.

A-Level Maths B6-11: Introducing the Factor TheoremWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Understand polynomial vocabulary — degree, coefficients, terms
State and apply the factor theorem: $f (p) = 0 \Leftrightarrow (x - p)$ is a factor
Use algebraic (long) division to remove a known factor
Fully factorise and solve a cubic

What a polynomial is

A $polynomial$ is a sum of terms of the form (number) $\times x^{whole number}$ , such as $2 x^{3} - 5 x^{2} + x - 7$ . The highest power is the $degree$ : degree 2 is a quadratic, degree 3 a cubic, degree 4 a quartic. Powers must be non-negative whole numbers — no roots or negative indices allowed.

A degree- $n$ polynomial has at most $n$ real roots, which is why a cubic can cross the $x$ -axis up to three times. Factorising is how you find those roots exactly.

The factor theorem

The factor theorem links roots and factors directly: for a polynomial $f (x)$ , the bracket $(x - p)$ is a factor $if and only if$ $f (p) = 0$ . In words — if putting $x = p$ into the polynomial gives zero, then $(x - p)$ divides it exactly.

This is why you $test values$ : try $x = 1, - 1, 2, - 2, \dots$ (usually factors of the constant term) until $f$ comes out zero. The mirror idea, the $remainder theorem$ , says $f (p)$ is exactly the remainder when you divide by $(x - p)$ — so a zero remainder means a clean factor.

f (p) = 0 ⟺ (x - p) is a factor of f (x)

A root and a factor are the same fact — watch the sign: the factor is

(x - p)

, and you test

x = + p

1The factor

(x - 2)

corresponds to testing

x = 2

2Substitute:

f (2) = 8 - 24 + 22 - 6 = 0

3Since

f (2) = 0

, by the factor theorem

(x - 2)

is a factor.

Answer

f (2) = 0

, so

(x - 2)

is a factor.

■

Tip — Mind the sign: the factor $(x - p)$ pairs with the value $x = p$ . So $(x + 3)$ is a factor when $f (- 3) = 0$ .

Algebraic division: removing the factor

Once you know one factor, divide it out to reduce the degree. Dividing a cubic by a linear factor leaves a quadratic, which you can then factorise with the usual tools. Polynomial long division works just like numerical long division — divide the leading terms, multiply back, subtract, and bring down.

1Divide

x^{3} - 6 x^{2} + 11 x - 6

(x - 2)

to get the quadratic

x^{2} - 4 x + 3

2Factorise the quadratic:

x^{2} - 4 x + 3 = (x - 1) (x - 3)

3Combine:

x^{3} - 6 x^{2} + 11 x - 6 = (x - 2) (x - 1) (x - 3)

Answer

(x - 1) (x - 2) (x - 3)

, so the roots are

x = 1, 2, 3

Why this works

Each factor you remove drops the degree by one, so a cubic becomes a quadratic you already know how to crack. Factorising is a chain: find one root by testing, divide it out, finish with quadratic methods.

Think like an examiner

To “show $(x - p)$ is a factor”, compute $f (p)$ and state explicitly that it equals 0 — the conclusion sentence earns a mark.
When testing for a root, try factors of the constant term first ( $\pm 1, \pm 2, \pm 3, \dots$ ) — the nice root is almost always there.
Set out algebraic division clearly; method marks are awarded for the process even if an arithmetic slip creeps in.
After factorising, answer the actual question — “solve” wants the roots, “factorise” wants the product of brackets.

Common misconceptions

Testing

x = - 2

to check the factor

(x - 2)

The factor

(x - 2)

pairs with

x = + 2

. Test

x = p

for the factor

(x - p)

Concluding

(x - p)

is a factor because the remainder is small.

It is a factor only when the remainder — i.e.

f (p)

— is exactly 0.

Stopping after finding one factor.

Divide it out and factorise the resulting quadratic to get all the roots.

Losing terms during long division by not lining up powers.

Keep columns aligned (include zero coefficients for missing powers) so nothing is dropped.

Factor theorem essentials

f (p) = 0 \Leftrightarrow (x - p) is a factor

The factor theorem.

f (p) = remainder on dividing by (x - p)

The remainder theorem.

de g (cubic) = 3 \Rightarrow up to 3 roots

Degree bounds the roots.

Stretch yourself

The polynomial $f (x) = 2 x^{3} + a x^{2} - 5 x - 6$ has $(x - 2)$ as a factor. Find $a$ , then factorise $f (x)$ fully.

Hint — Use $f (2) = 0$ to form an equation for $a$ . Then divide by $(x - 2)$ and factorise the quadratic.

Questions students ask

What is the factor theorem in simple terms?

If substituting

x = p

into a polynomial gives zero, then

(x - p)

divides it exactly — it is a factor. It lets you find a factor by testing values rather than guessing brackets.

How do I know which values to test?

Try the factors of the constant term first: for

x^{3} - 6 x^{2} + 11 x - 6

, test

\pm 1, \pm 2, \pm 3, \pm 6

. A whole-number root, if there is one, is almost always among them.

What is the difference between the factor and remainder theorems?

The remainder theorem says

f (p)

is the remainder when you divide by

(x - p)

. The factor theorem is the special case where that remainder is zero — meaning

(x - p)

is a factor.

Do I have to use long division?

You need some way to remove a known factor — long division is the standard method, though comparing coefficients (writing

f (x) = (x - p) (quadratic)

and matching) works too. Either is fine as long as it is clearly set out.

Key takeaways

A polynomial’s degree is its highest power; a degree- $n$ polynomial has at most $n$ roots.
Factor theorem: $(x - p)$ is a factor exactly when $f (p) = 0$ — so test $x = p$ .
Watch the sign: $(x + 3)$ is a factor when $f (- 3) = 0$ .
Divide out a known factor to reduce the degree, then finish with quadratic methods.
Not every cubic has three real roots — check the leftover quadratic’s discriminant.

How this fits the course

Build on

Partial Fractions

Leads to

Graphs and Transformations

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