CDS Mathematics · Polynomials

Factorisation of Polynomials

Factorise a cubic by finding one root and dividing, a product of four brackets by pairing them, and everything else by spotting a standard identity.

All Polynomials notes CDS Mathematics strategy

Why this matters

Nineteen PYQs. Three routes cover them: find a small integer root and divide; pair four linear brackets so a common quadratic appears and substitute; or recognise a known identity — difference of squares or cubes, a perfect square, grouping.

Concept 1 of 3: Cubics: find a root, then divide

An integer root of a monic cubic must divide the constant term. Try the small divisors, and the first one that gives zero splits off a linear factor; the quadratic left over factorises by the usual method.

Definition

Integer roots of $x^3 + bx^2 + cx + d$ divide $d$ : try $\pm 1, \pm 2, \ldots$ .
Once $x - a$ is found, divide (or compare coefficients) to get the quadratic.
A monic quartic with a known cubic factor $(x - 1)^3$ has its last factor fixed by the constant term.
Grouping: $x^3(x^2 + 2x + 1) - (x^2 + 2x + 1) = (x^3 - 1)(x + 1)^2$ .

Rational root test (monic)

x^3 + bx^2 + cx + d:\ \text{integer roots divide } d

Worked example

Factorise

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (II) 2017 — Elementary Mathematics · Q3Easy

Example 1 · Polynomials · Factorisation of Polynomials

What are the factors of

x^3 + 4x^2 - 11x - 30

Check the sign of each root

f(-2) = 0

, the factor is

x + 2

, not

x - 2

. Options are built from the same numbers with the signs changed, so verify one root in the original.

Drill 5 more on cubics: find a root, then divide

Concept 2 of 3: Pair the brackets and substitute

x(x + 2)(x + 3)(x + 5)

, pairing the outer and inner brackets gives

x^2 + 5x

and

x^2 + 5x + 6

— the same quadratic plus a constant. Call it

t

and the expression becomes a quadratic in

t

Definition

Pair brackets so that each pair has the same $x^2 + bx$ part: the constants in the brackets must add to the same total in each pair.
Substitute $t = x^2 + bx$ , factorise in $t$ , then substitute back.
Expressions in $(3x + y)$ and $(x + 5y)$ : put $A = 3x + y$ , $B = x + 5y$ and factorise in $A, B$ .
$x(x + 1)(x + 2)(x + 3) + 1 = (x^2 + 3x + 1)^2$ .

Pairing four brackets

x(x + 3)\cdot(x + 1)(x + 2) = t(t + 2), \quad t = x^2 + 3x

Worked example

Factorise

(x + 1)(x + 2)(x + 3)(x + 4) - 24

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — Elementary Mathematics · Q19Moderate

Example 2 · Polynomials · Factorisation of Polynomials

x(x - 1)(x - 2)(x - 3) + 1 = k^2

, then which one of the following is a possible expression for

k

Pair by equal sums

x(x + 2)(x + 3)(x + 5)

, pair

0

with

5

and

2

with

3

(both sum to

5

). Pairing

x

with

x + 2

gives two different quadratics and no substitution.

Drill 4 more on pair the brackets and substitute

Concept 3 of 3: Factorising with identities

Many exam polynomials are a standard identity in disguise: a difference of squares or cubes, a perfect square, or a sum that groups into a common bracket. Recognising the shape is the whole solution.

Definition

$a^2 - b^2$ , $a^3 \pm b^3$ , $a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$ .
Perfect square: $(px^2 + qx + r)^2$ — check the first term, the last term and the cross term $2pr$ .
Grouping: $1 - x - x^n + x^{n + 1} = (1 - x)(1 - x^n)$ , and $1 - x^n$ has the factor $1 - x$ .
$u^3 + v^3 + w^3 - 3uvw = (u + v + w)(u^2 + v^2 + w^2 - uv - vw - wu)$ .

A useful quartic

a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)

Worked example

Find the square root of

9x^4 - 12x^3 + 10x^2 - 4x + 1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q39Moderate

Example 3 · Polynomials · Factorisation of Polynomials

1 - x - x^n + x^{n+1}

, where n is a natural number, is divisible by

Plus or minus in the linear factor

u^3 + v^3 + w^3 - 3uvw

has the factor

u + v + w

with every sign PLUS. An option with one term negated is not a factor, even though it looks close.

Drill 7 more on factorising with identities

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)

Cubics: find a root, then divide
Rational root test (monic)
$x^3 + bx^2 + cx + d:\ \text{integer roots divide } d$
Pair the brackets and substitute
Pairing four brackets
$x(x + 3)\cdot(x + 1)(x + 2) = t(t + 2), \quad t = x^2 + 3x$
Factorising with identities
A useful quartic
$a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$

Watch out for (3)

Check the sign of each root
→ Cubics: find a root, then divide
Pair by equal sums
→ Pair the brackets and substitute
Plus or minus in the linear factor
→ Factorising with identities

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 19 questions