Polynomials

What you'll learn

Define polynomials and identify their degree, coefficients, and types
Apply the Remainder Theorem to find remainders without long division
Use the Factor Theorem to test if a binomial is a factor
State and use five standard algebraic identities
Factorise cubic and quadratic polynomials using identities

Key concepts

Definitions — Degree, Coefficients, Types

A polynomial in variable x is an expression of the form: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀

where n is a non-negative integer and aₙ ≠ 0 is the leading coefficient.

Key terms:

Term	Definition	Example in 3x³ − 5x + 2
Degree	Highest power of variable	3
Leading coefficient	Coefficient of highest-degree term	3
Constant term	Term with no variable (a₀)	2
Coefficient of x	Number multiplied by x	−5

Types by degree:

Name	Degree	Example
Constant polynomial	0	7
Linear polynomial	1	2x + 3
Quadratic polynomial	2	x² − 4x + 1
Cubic polynomial	3	2x³ + x − 5
Biquadratic	4	x⁴ − 3x² + 2

Types by number of terms:

Name	Terms	Example
Monomial	1	5x²
Binomial	2	x³ − 4
Trinomial	3	x² + 3x − 2

Zero of a polynomial p(x): A value 'a' such that p(a) = 0.

Worked Example: Find the zero of p(x) = 3x + 6. 3x + 6 = 0 → x = −2. So zero = −2.

Note: A polynomial of degree n has at most n zeros.

Remainder Theorem

Statement: If a polynomial p(x) is divided by (x − a), the remainder is p(a).

Worked Example 1: Find the remainder when p(x) = x³ − 3x² + 5x − 2 is divided by (x − 2). p(2) = 8 − 12 + 10 − 2 = 4 Remainder = 4 (no long division needed!)

Worked Example 2: Find the remainder when 2x³ + x − 3 is divided by (x + 1). Divisor = x − (−1), so a = −1 p(−1) = 2(−1)³ + (−1) − 3 = −2 − 1 − 3 = −6

Worked Example 3: Find the value of k if (x + 2) leaves remainder 0 when dividing x³ + kx² + x − 6. a = −2: p(−2) = −8 + 4k − 2 − 6 = 0 → 4k − 16 = 0 → k = 4

Factor Theorem

Statement: (x − a) is a factor of polynomial p(x) if and only if p(a) = 0.

Factor theorem = special case of remainder theorem where remainder = 0.

Worked Example 1: Is (x − 3) a factor of p(x) = x³ − 4x² + x + 6? p(3) = 27 − 36 + 3 + 6 = 0 ✓ → Yes, (x − 3) is a factor

Worked Example 2: Is (x + 2) a factor of x³ + 3x² − x − 3? p(−2) = −8 + 12 + 2 − 3 = 3 ≠ 0 → No, (x + 2) is NOT a factor

Using factor theorem to fully factorise a cubic:

Worked Example: Factorise x³ − 2x² − 5x + 6. Try x = 1: 1 − 2 − 5 + 6 = 0 ✓ → (x − 1) is a factor. Divide: x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6) Factorise x² − x − 6 = (x − 3)(x + 2) Final: (x − 1)(x − 3)(x + 2)

Algebraic Identities

Five standard identities at Class 9 level:

#	Identity
1	(x + y)² = x² + 2xy + y²
2	(x − y)² = x² − 2xy + y²
3	x² − y² = (x + y)(x − y)
4	(x + a)(x + b) = x² + (a+b)x + ab
5	(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
6	(x + y)³ = x³ + 3x²y + 3xy² + y³ = x³ + y³ + 3xy(x + y)
7	(x − y)³ = x³ − 3x²y + 3xy² − y³ = x³ − y³ − 3xy(x − y)
8	x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)

Key derived results:

x³ + y³ = (x + y)(x² − xy + y²)
x³ − y³ = (x − y)(x² + xy + y²)
If x + y + z = 0, then x³ + y³ + z³ = 3xyz

Using Identity 4:

(x + 3)(x − 5) = x² + (3 + (−5))x + (3 × (−5)) = x² − 2x − 15

Using Identity 5:

(a + b + c)² where a=2, b=3, c=−1: = 4 + 9 + 1 + 2(2)(3) + 2(3)(−1) + 2(2)(−1) = 14 + 12 − 6 − 4 = 16 ✓ Direct: (2+3−1)² = 4² = 16 ✓

Factorisation Using Identities

Quadratic factorisation (Identity 3 and 4):

x² − 25 = x² − 5² = (x+5)(x−5)

x² + 7x + 12 = x² + (3+4)x + (3)(4) = (x+3)(x+4)

x² − x − 6 = x² + (2+(−3))x + (2)(−3) = (x+2)(x−3)

Sum and difference of cubes:

8x³ + 27 = (2x)³ + 3³ = (2x + 3)((2x)² − (2x)(3) + 9) = (2x+3)(4x² − 6x + 9)

125a³ − 8b³ = (5a)³ − (2b)³ = (5a − 2b)(25a² + 10ab + 4b²)

Identity 8 application:

a³ + b³ + c³ − 3abc when a+b+c ≠ 0: = (a+b+c)(a²+b²+c²−ab−bc−ca)

If a + b + c = 0 → a³ + b³ + c³ = 3abc

Worked Example: Find 1³ + (−1)³ + 0³. a + b + c = 1 + (−1) + 0 = 0 So 1³ + (−1)³ + 0³ = 3(1)(−1)(0) = 0 ✓

Complete factorisation strategy:

Pattern detected	Identity/Method	Result
a² − b²	Identity 3	(a+b)(a−b)
a² + 2ab + b²	Identity 1	(a+b)²
a² − 2ab + b²	Identity 2	(a−b)²
x² + (a+b)x + ab	Identity 4	(x+a)(x+b)
a³ + b³	Sum of cubes	(a+b)(a²−ab+b²)
a³ − b³	Difference of cubes	(a−b)(a²+ab+b²)
Cubic with known zero	Factor theorem	Long divide, then factorise quotient

Quick check

Find the degree and number of zeros (maximum) of p(x) = 2x⁴ − x³ + 3x − 7.
Find the remainder when 3x³ − 4x² + 7 is divided by (x + 2) using the Remainder Theorem.
Show that (x − 2) is a factor of x³ − 6x² + 11x − 6, then fully factorise.
Expand (2a − 3b + c)² using Identity 5.
Factorise: 27x³ − 125y³

Open the Practice tab for graded questions on Polynomials.