Algebra

What you'll learn

Classify terms in algebraic expressions by type and degree
Multiply monomials, binomials, and polynomials
Apply and derive the three standard algebraic identities
Factorise expressions using common factors, grouping, and identities

Key concepts

Algebraic Expressions — Terms, Like/Unlike, Degree

Parts of an algebraic expression:

For 5x³ − 3x²y + 2xy − 7:

Term	Coefficient	Variables	Degree of term
5x³	5	x	3
−3x²y	−3	x, y	3 (2+1)
2xy	2	x, y	2 (1+1)
−7	−7	none	0

Degree of a term = sum of exponents of all variables in that term. Degree of expression = highest degree among all terms = 3 in the example above.

Like vs Unlike terms:

Like (can be combined)	Unlike (cannot be combined)
4x² and −7x²	4x² and 4x
3xy and −xy	3xy and 3x²y
5 and −2 (constants)	5 and 5x

Addition/Subtraction: Combine like terms only.

Worked Example: (3x² + 5xy − 2y) + (−x² + 3xy + 4y) = (3−1)x² + (5+3)xy + (−2+4)y = 2x² + 8xy + 2y

Multiplication of Algebraic Expressions

Monomial × Monomial

Multiply coefficients and add exponents of same variables.

3x² × (−4xy) = −12x³y (−5a²b) × (2ab³) = −10a³b⁴

Monomial × Polynomial

Distribute the monomial to every term.

3x × (2x² − 4x + 5) = 6x³ − 12x² + 15x

Binomial × Binomial (FOIL / horizontal method)

(a + b)(c + d) = ac + ad + bc + bd

Worked Example: (2x + 3)(x − 4) = 2x·x + 2x·(−4) + 3·x + 3·(−4) = 2x² − 8x + 3x − 12 = 2x² − 5x − 12

Polynomial × Polynomial

Multiply each term of the first polynomial with every term of the second.

Worked Example: (x² + 2x − 1)(x + 3) = x²·x + x²·3 + 2x·x + 2x·3 + (−1)·x + (−1)·3 = x³ + 3x² + 2x² + 6x − x − 3 = x³ + 5x² + 5x − 3

Algebraic Identities

An identity is an equation true for all values of the variables.

Identity 1: (a + b)² = a² + 2ab + b²

Derivation: (a + b)² = (a + b)(a + b) = a² + ab + ba + b² = a² + 2ab + b²

Worked Example: Expand (3x + 5)² a = 3x, b = 5 = (3x)² + 2(3x)(5) + 5² = 9x² + 30x + 25

Numerical use: 102² = (100 + 2)² = 10000 + 400 + 4 = 10404

Identity 2: (a − b)² = a² − 2ab + b²

Worked Example: Expand (4y − 3)² a = 4y, b = 3 = 16y² − 24y + 9

Numerical use: 98² = (100 − 2)² = 10000 − 400 + 4 = 9604

Identity 3: (a + b)(a − b) = a² − b²

Proof: (a+b)(a−b) = a² − ab + ab − b² = a² − b²

Worked Example: (5x + 7)(5x − 7) = (5x)² − 7² = 25x² − 49

Numerical use: 105 × 95 = (100+5)(100−5) = 10000 − 25 = 9975

Summary of identities:

Identity	Formula
Square of sum	(a+b)² = a² + 2ab + b²
Square of difference	(a−b)² = a² − 2ab + b²
Difference of squares	(a+b)(a−b) = a² − b²
Relationship	(a+b)² − (a−b)² = 4ab
Relationship	(a+b)² + (a−b)² = 2(a² + b²)

Factorisation

Factorisation is the reverse of expansion — expressing a polynomial as a product of factors.

Method 1: Taking Out Common Factors (HCF Method)

Find the HCF of all terms, factor it out.

6x²y + 9xy² − 3xy HCF = 3xy = 3xy(2x + 3y − 1)

Method 2: Regrouping

Rearrange terms, take common factors from groups.

4xy + 2x + 6y + 3 = 2x(2y + 1) + 3(2y + 1) = (2y + 1)(2x + 3)

ax − ay + bx − by = a(x − y) + b(x − y) = (x − y)(a + b)

Method 3: Using Identity 1 (a + b)²

Recognise a² + 2ab + b² as (a + b)²

x² + 10x + 25 = x² + 2(x)(5) + 5² = (x + 5)²

4a² + 12ab + 9b² = (2a)² + 2(2a)(3b) + (3b)² = (2a + 3b)²

Method 4: Using Identity 2 (a − b)²

Recognise a² − 2ab + b² as (a − b)²

9x² − 24x + 16 = (3x)² − 2(3x)(4) + 4² = (3x − 4)²

Method 5: Using Identity 3 (a² − b²)

Recognise a² − b² as (a + b)(a − b)

25x² − 49 = (5x)² − 7² = (5x + 7)(5x − 7)

x⁴ − 1 = (x²)² − 1² = (x² + 1)(x² − 1) = (x² + 1)(x + 1)(x − 1)

Comparison table — factorisation strategies:

Pattern	Method	Factor form
Each term has common factor	HCF	HCF × (remaining)
4 terms, pair-able	Grouping	( )( )
a² + 2ab + b²	Identity 1	(a + b)²
a² − 2ab + b²	Identity 2	(a − b)²
a² − b²	Identity 3	(a+b)(a−b)

Quick check

Find the degree of: 5x²y³ − 3x⁴ + 7xy − 9
Expand using identity: (3a − 2b)² and verify by direct multiplication.
Factorise: 12a²b − 18ab² + 24ab
Factorise: x² − 81
Factorise by grouping: 6xy − 4y + 3x − 2

Open the Practice tab for graded questions on Algebra.