Algebraic processes underpin almost everything in the Edexcel IGCSE (4MB1) Mathematics Specification B specification. Whether you are solving equations, working with functions, or tackling calculus, the algebraic toolkit you build here is what makes those topics accessible.
Key facts
- Algebraic processes is among the most commonly examined areas across both Paper 1 and Paper 2.
- The topic covers collecting like terms, the four operations on algebraic expressions, index laws, formulae, factorisation (including the factor theorem), algebraic division, and manipulation of algebraic fractions.
- Questions range from 1-mark simplifications to multi-step problems worth 5 or 6 marks.
- The edexcel igcse algebraic processes content requires fluency, not just understanding. Speed and accuracy matter.
Collecting like terms and the four operations
The foundation of algebra is simplifying expressions by combining terms that share the same variable and power. Terms are "like" if they have the same variable raised to the same power.
Worked example: Simplify 3x2 + 5x - 2x2 + 7 - 3x + 1
Group by type: (3x2 - 2x2) + (5x - 3x) + (7 + 1)
= x2 + 2x + 8
For multiplication and division, apply index laws. When multiplying terms with the same base, add the indices. When dividing, subtract them.
Worked example: Simplify (2x3)(5x4)
Multiply coefficients: 2 x 5 = 10
Add indices: 3 + 4 = 7
Result: 10x7
Index laws with integer, fractional and negative powers
The algebraic processes explained here extend well beyond positive integer powers. You need all of these rules, and the edexcel igcse mathematics specification b explained approach is to master each one individually before combining them.
| Rule | Expression | Example |
|---|---|---|
| Multiplication | am x an = am+n | x3 x x5 = x8 |
| Division | am / an = am-n | x7 / x2 = x5 |
| Power of a power | (am)n = amn | (x3)4 = x12 |
| Negative index | a-n = 1/an | x-2 = 1/x2 |
| Fractional index | a1/n = nth root of a | 271/3 = 3 |
| Zero index | a0 = 1 | 50 = 1 |
Worked example: Simplify (16x8)3/4
Apply the power to both the coefficient and the variable:
163/4 = (161/4)3 = 23 = 8
(x8)3/4 = x8 x 3/4 = x6
Result: 8x6
Formulae: construction, substitution and change of subject
You must be able to build a formula from a description, substitute values into a formula, and rearrange a formula to make a different variable the subject.
Worked example (change of subject): Make r the subject of A = pi r2
A = pi r2
A / pi = r2
r = sqrt(A / pi)
Worked example (with fractions): Make x the subject of y = (2x + 1) / (x - 3)
Multiply both sides by (x - 3): y(x - 3) = 2x + 1
Expand: yx - 3y = 2x + 1
Collect x terms: yx - 2x = 3y + 1
Factor out x: x(y - 2) = 3y + 1
x = (3y + 1) / (y - 2)
Factorisation
There are several types of factorisation the exam tests, from basic common-factor extraction to the factor theorem for cubics.
Common factor
6x2 + 9x = 3x(2x + 3)
Difference of two squares
x2 - 25 = (x + 5)(x - 5)
Quadratic trinomial
x2 + 7x + 12 = (x + 3)(x + 4)
Harder quadratic (coefficient of x2 greater than 1)
6x2 + x - 2
Find two numbers that multiply to 6 x (-2) = -12 and add to 1. Those numbers are 4 and -3.
Rewrite: 6x2 + 4x - 3x - 2
Group: 2x(3x + 2) - 1(3x + 2)
= (2x - 1)(3x + 2)
The factor theorem
The factor theorem states: if f(a) = 0, then (x - a) is a factor of f(x). For factors of the form (ax + b), if f(-b/a) = 0, then (ax + b) is a factor. This is a standard part of the what is algebraic processes igcse specification.
Worked example: Show that (x - 2) is a factor of f(x) = x3 - 5x2 + 2x + 8.
Substitute x = 2:
f(2) = 8 - 20 + 4 + 8 = 0
Since f(2) = 0, (x - 2) is a factor.
Algebraic division of a cubic by a linear factor
Once you have found one factor using the factor theorem, you can divide the cubic by that factor to find the remaining quadratic, which you then factorise further.
Worked example: Factorise f(x) = x3 - 5x2 + 2x + 8 completely.
From above, (x - 2) is a factor. Divide x3 - 5x2 + 2x + 8 by (x - 2).
Using polynomial long division or comparison of coefficients:
x3 - 5x2 + 2x + 8 = (x - 2)(x2 - 3x - 4)
Now factorise x2 - 3x - 4 = (x - 4)(x + 1)
Complete factorisation: (x - 2)(x - 4)(x + 1)
Check: the roots are x = 2, 4 and -1. Substituting each into the original cubic gives 0. Confirmed.
Algebraic fractions
The specification requires you to add, subtract, multiply and divide algebraic fractions with numerical, linear or quadratic denominators. The key skill is finding a common denominator.
Worked example (addition): Simplify 3/(x + 1) + 2/(x - 3)
Common denominator: (x + 1)(x - 3)
= 3(x - 3)/[(x + 1)(x - 3)] + 2(x + 1)/[(x + 1)(x - 3)]
= [3(x - 3) + 2(x + 1)] / [(x + 1)(x - 3)]
= [3x - 9 + 2x + 2] / [(x + 1)(x - 3)]
= (5x - 7) / [(x + 1)(x - 3)]
Worked example (simplification): Simplify (x2 - 9) / (x2 + 5x + 6)
Factorise numerator: (x + 3)(x - 3)
Factorise denominator: (x + 2)(x + 3)
Cancel (x + 3): (x - 3) / (x + 2)
Exam question patterns
The edexcel igcse mathematics specification b definition of algebraic processes covers a wide range, and the exam tests it in several recognisable patterns.
- "Simplify" questions: Reduce an expression to its simplest form. Common with algebraic fractions and index expressions.
- "Factorise completely" questions: Extract all possible factors. "Completely" means do not stop at a partial factorisation.
- "Show that" with algebra: Prove an algebraic identity by expanding, simplifying or rearranging.
- "Make x the subject" questions: Rearrange a formula. These can involve squaring, square-rooting, or dealing with x in the denominator.
- "Use the factor theorem" questions: Test whether a given expression is a factor of a polynomial, then complete the factorisation.
Self-check questions
Work through each one before checking the answer.
- Simplify (3x4y2) x (4x-2y3)
Answer: 12x2y5 - Factorise completely: 2x3 - 18x
Answer: 2x(x2 - 9) = 2x(x + 3)(x - 3) - Make t the subject of v = u + at
Answer: t = (v - u) / a - Simplify: 2/(x - 1) - 3/(x + 2)
Answer: [2(x + 2) - 3(x - 1)] / [(x - 1)(x + 2)] = (2x + 4 - 3x + 3) / [(x - 1)(x + 2)] = (-x + 7) / [(x - 1)(x + 2)] = (7 - x) / [(x - 1)(x + 2)] - Use the factor theorem to show that (x + 1) is a factor of x3 + 3x2 - x - 3.
Answer: f(-1) = -1 + 3 + 1 - 3 = 0. Since f(-1) = 0, (x + 1) is a factor.
Expanding double brackets and higher
Expanding two brackets is tested regularly, and the method extends to expanding a bracket multiplied by a quadratic. For double brackets, multiply every term in the first bracket by every term in the second.
Worked example: Expand (2x + 3)(x - 5)
= 2x(x) + 2x(-5) + 3(x) + 3(-5)
= 2x2 - 10x + 3x - 15
= 2x2 - 7x - 15
Worked example: Expand (x + 2)(x2 - 3x + 4)
= x(x2) + x(-3x) + x(4) + 2(x2) + 2(-3x) + 2(4)
= x3 - 3x2 + 4x + 2x2 - 6x + 8
= x3 - x2 - 2x + 8
Write each product separately before collecting like terms. This keeps the working transparent and earns method marks even if the final simplification contains a slip.
Substitution into formulae
Substitution questions require you to replace variables with given values and evaluate the expression. The most common errors are sign mistakes and incorrect order of operations.
Worked example: Given that v = u + at, find v when u = 5, a = -3 and t = 4.
v = 5 + (-3)(4) = 5 + (-12) = 5 - 12 = -7
Always use brackets around negative values to prevent sign confusion. Write (-3)(4) rather than -3 x 4, especially when substituting into more complex expressions.
Algebraic processes is a topic where consistent practice makes a visible difference. The edexcel igcse mathematics specification b notes above cover every assessable skill, but reading them is not enough. Work through problems until the methods feel automatic, and use the Green Bridge CBT platform for edexcel igcse mathematics specification b practice questions that mirror real exam challenges.
Edexcel IGCSE algebraic processes explained: collecting terms, index laws, factorisation, the factor theorem, algebraic division and algebraic fractions.
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