Algebraic Manipulation & Indices

Mathematics· 2.1, 2.2, 2.4 (2025–2027 syllabus)· 25 min read

1. Substitution & Simplifying Like Terms★★☆☆☆⏱ 5 min

Substitution is the process of replacing variables in an expression or formula with given numerical values, then simplifying using BODMAS order of operations. To simplify expressions, you first group and combine like terms.

📐 Worked Example

Evaluate $2 x^{2} - 3 y + 5$ when $x = 2$ and $y = - 1$ .

1
Substitute the given values for x and y
$2(2)^2 - 3(-1) + 5$
2
Calculate exponents first, then multiply
$2(4) + 3 + 5 = 8 + 3 +5$
3
Add terms to get the final result
$16$

📘 Definition

Like Terms

Terms that have identical variable parts, e.g. $3 x$ and $- 7 x$ are like terms, but $2 x^{2}$ and $4 x$ are not.

Example:

Simplify $3 a + 2 b - 5 a + 7 b$

📐 Worked Example

Simplify $3 a + 2 b - 5 a + 7 b$ .

1
Group like terms together, keeping their signs
$(3a - 5a) + (2b + 7b)$
2
Combine coefficients of each group
$-2a +9b$

Exam tip:

Always carry the sign of each term when collecting like terms: negative signs apply only to the term immediately after them.

2. Expanding Brackets (Core)★★☆☆☆⏱ 5 min

Expanding brackets uses the distributive law: every term inside the bracket is multiplied by the term outside the bracket. For products of two brackets, multiply every term in the first bracket by every term in the second.

📐 Worked Example

Expand $4 (2 x - 3 y)$ .

1
Multiply each term inside the bracket by 4
$4 \times 2x - 4 \times 3y$
2
Simplify to get the result
$8x - 12y$

📐 Worked Example

Expand $(x + 3) (2 x - 5)$ .

1
Apply FOIL rule
$x \times 2x + x \times (-5) + 3 \times 2x + 3 \times (-5)$
2
Calculate each product
$2x^2 -5x +6x -15$
3
Combine like terms
$2x^2 +x -15$

Exam tip:

After expanding pairs of brackets, always simplify by collecting like terms to get full marks.

3. Core Factorisation & Integer Index Laws★★★☆☆⏱ 6 min

Core factorisation involves extracting the highest common factor (HCF) of all terms in an expression. Integer index laws are used to simplify expressions with powers of the same base.

📘 Definition

Highest Common Factor (HCF) for algebra

The largest number and highest power of each variable that divides all terms in the expression.

Example:

HCF of $6 x^{2}$ and $9 x$ is $3 x$

📐 Worked Example

Factorise $12 x^{3} y - 8 x^{2} y^{2}$ .

1
Identify HCF of all terms: 4, $x^{2}$ , $y$ → HCF = $4 x^{2} y$
2
Divide each term by the HCF and write the result in brackets
$4x^2y(3x - 2y)$

Multiplication: $a^{m} \times a^{n} = a^{m + n}$
Division: $a^{m} \div a^{n} = a^{m - n}$
Power of a power: $(a^{m})^{n} = a^{mn}$
Zero index: $a^{0} = 1$ for $a \neq = 0$
Negative index: $a^{- n} = \frac{1}{a ^{n}}$ for $a \neq = 0$

📐 Worked Example

Simplify $\frac{15 a ^{4} b ^{2}}{3 a ^{2} b} \times (2 a^{3})^{2}$ .

1
Simplify the fraction first using division law
$5a^{2}b \times (2a^3)^2$
2
Apply power of a power law to the second term
$5a^2b \times 4a^6$
3
Multiply coefficients and apply multiplication law for indices
$20a^8b$

📐 Worked Example

Solve $3^{2 x - 1} = 27$ for $x$ .

1
Rewrite both sides with the same base: $27 = 3^{3}$
$3^{2x-1} = 3^3$
2
Equate exponents since the bases are equal
$2x - 1 = 3$
3
Solve for x
$2x = 4 → x = 2$

Exam tip:

Always verify your factorised answer by expanding it back out to match the original expression.

4. Extended Adds: Advanced Expansion & Factorisation★★★★☆Extended only⏱ 5 min

Extended tier requires expanding products of 3+ brackets, and factorising using 5 special forms: common factor, difference of two squares (DOTS), grouping, $a x^{2} + b x + c$ , and $a^{2} x^{2} - b^{2} y^{2}$ .

📐 Worked Example

Expand $(x + 2) (x - 3) (2 x + 1)$ .

1
Expand first two brackets first
$(x^2 - 3x +2x -6)(2x +1) = (x^2 -x -6)(2x+1)$
2
Multiply each term in the first bracket by each term in the second
$x^2(2x+1) -x(2x+1) -6(2x+1)$
3
Simplify and collect like terms
$2x^3 +x^2 -2x^2 -x -12x -6 = 2x^3 -x^2 -13x -6$

📐 Worked Example

Factorise $2 x^{2} - 7 x - 15$ .

1
Find factors of $2 x^{2}$ (2x and x) and factors of -15 that sum to -7x when cross-multiplied: +3 and -5
2
Write the factorised form
$(2x +3)(x -5)$
3
Verify by expanding to match original expression

Exam tip:

For quadratic factorisation with $a > 1$ , test pairs of factors for the $x^{2}$ coefficient and constant term until you get the correct middle term when expanded.

5. Extended Adds: Completing the Square & Advanced Indices★★★★☆Extended only⏱ 4 min

Extended additionally requires completing the square for quadratics and applying fractional index laws such as $a^{m / n} = (n a)^{m}$ . (Negative indices and simple same-base index equations are Core content, covered above.)

📐 Worked Example

Complete the square for $2 x^{2} + 8 x - 3$ .

1
Factor out the coefficient of $x^{2}$ from the first two terms
$2(x^2 +4x) -3$
2
Complete the square inside the bracket: $x^{2} + 4 x = (x + 2)^{2} - 4$
3
Substitute back and simplify the constant term
$2[(x+2)^2 -4] -3 = 2(x+2)^2 -8 -3 = 2(x+2)^2 -11$

Fractional index: $a^{1/ n} = n a$ , $a^{m / n} = (n a)^{m}$

📐 Worked Example

Evaluate $1 6^{3/4}$ .

1
Take the $n$ th root first: $1 6^{1/4} = 416 = 2$
$16^{1/4} = 2$
2
Raise to the numerator power
$16^{3/4} = 2^3 = 8$

6. Common Pitfalls

Wrong move:

Forgetting to apply a negative coefficient to all terms when expanding a bracket

Why:

Negative numbers multiply through every term inside the bracket, not just the first term

Correct move:

Multiply every term inside the bracket by the full coefficient (including its sign) before simplifying

Wrong move:

Adding exponents when multiplying terms with different bases

Why:

Index laws only apply to terms with identical bases

Correct move:

Keep terms with different bases separate when simplifying index expressions

Wrong move:

Trying to factorise a sum of two squares (e.g. $x^{2} + 4$ ) using DOTS rule

Why:

Sum of two squares has no linear factors at IGCSE level

Correct move:

Only apply DOTS to expressions of the form $a^{2} - b^{2}$

Wrong move:

Forgetting to adjust the constant term after factoring out the leading coefficient when completing the square

Why:

Factoring changes the value of the expression, so you need to compensate to keep it equivalent

Correct move:

After completing the square inside the factored bracket, expand and adjust the constant term to match the original expression

Wrong move:

Treating $a^{0}$ as 0 instead of 1 for non-zero $a$

Why:

Any non-zero value raised to the power of 0 equals 1 by definition of index laws

Correct move:

Immediately replace any non-zero base raised to power 0 with 1 when simplifying

7. Quick Reference Cheatsheet

Skill	Tier	Rule/Method
Substitution	Core	Replace variables with given values, follow BODMAS order
Collect like terms	Core	Group terms with identical variable parts, combine coefficients
Expand single bracket	Core	$a (b + c) = ab + a c$ , apply sign of coefficient to all terms
Expand two brackets	Core	Use FOIL rule, then collect like terms to simplify
Core factorisation	Core	Extract the highest common factor from all terms
Index laws (positive, zero, negative)	Core	$a^{m} \times a^{n} = a^{m + n}$ , $a^{m} \div a^{n} = a^{m - n}$ , $(a^{m})^{n} = a^{mn}$ , $a^{0} = 1$ , $a^{- n} = 1/ a^{n}$
Expand 3+ brackets	Extended	Expand first two brackets first, multiply result by remaining bracket
Special factorisation	Extended	Common factor, DOTS, grouping, $a x^{2} + b x + c$ , $a^{2} x^{2} - b^{2} y^{2}$
Completing the square	Extended	Rewrite $a x^{2} + b x + c$ as $a (x + p)^{2} + q$
Fractional indices	Extended	$a^{1/ n} =^{n} \sqrt a$ , $a^{m / n} = (^{n} \sqrt a)^{m}$
Simple index equations	Core	Rewrite both sides with same base, equate exponents to solve

What's Next

Now that you have mastered algebraic manipulation and indices for CIE IGCSE Maths 0580, you are ready to apply these foundational skills to more advanced algebra topics. These skills are prerequisites for solving linear and quadratic equations, rearranging formulae, and working with algebraic fractions, all of which appear frequently on both Core and Extended papers. You will also use index rules when working with standard form, sequences, and exponential graphs later in the syllabus. For Extended students, your understanding of completing the square will directly support solving quadratic equations and identifying key features of quadratic graphs in upcoming units. Be sure to practice past paper questions regularly to reinforce your speed and accuracy with these skills, as they are tested in almost every exam paper.