What is an algebraic expression in the context of Edexcel International A Levels Pure Mathematics 1?

An algebraic expression in Edexcel International A Levels Pure Mathematics 1 is a mathematical phrase that can include numbers, variables, and operation symbols. It represents a value and can be as simple as a single number or variable, or more complex with multiple terms and operations. Understanding algebraic expressions is fundamental for solving equations and inequalities in Pure Mathematics.

How do you simplify algebraic expressions in Pure Mathematics 1?

To simplify algebraic expressions in Pure Mathematics 1, you need to combine like terms, which are terms that have the same variable raised to the same power. You can also use the distributive property to expand expressions and then combine like terms. Simplification helps in making expressions easier to work with, especially when solving equations or inequalities.

What are the common mistakes to avoid when working with algebraic expressions in Edexcel International A Levels?

Common mistakes include incorrectly combining unlike terms, misapplying the distributive property, and neglecting to apply the correct order of operations. It's important to carefully follow mathematical rules and double-check each step to ensure accuracy when manipulating algebraic expressions in Edexcel International A Levels.

How do you factorize algebraic expressions in Pure Mathematics 1?

Factorizing algebraic expressions involves expressing the expression as a product of its factors. This can be done by finding common factors, using the difference of squares, or applying the quadratic formula for quadratic expressions. Factorization is a key skill in Pure Mathematics 1, as it simplifies solving equations and understanding polynomial functions.

Why is understanding algebraic expressions important for Edexcel International A Levels Mathematics?

Understanding algebraic expressions is crucial for Edexcel International A Levels Mathematics because they form the foundation for more advanced topics such as equations, functions, and calculus. Mastery of algebraic expressions enables students to manipulate and solve mathematical problems effectively, which is essential for success in Pure Mathematics 1 and beyond.

Why is "Writing $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$" a common mistake in Algebraic Expressions?

Why it happens: Over-extending the product rule √(ab) = √(a)√(b) to sums. How to avoid it: Test with numbers: √(9 + 16) = √(25) = 5, NOT 3 + 4 = 7. The square root does NOT distribute over addition. Only over multiplication and division. Source: WMA11/01 examiner reports flag this regularly

Why is "Computing $16^{3/4}$ as $16 \div (3/4) = 21.33$ or as $12$" a common mistake in Algebraic Expressions?

Why it happens: Confusing index notation with division. How to avoid it: a^m/n means 'take the n-th root, then raise to the m-th power'. So 16^3/4 = (16^1/4)^3 = 2^3 = 8. Always take the root FIRST.

Why is "Treating $a^{-n}$ as $-a^{n}$" a common mistake in Algebraic Expressions?

Why it happens: Mixing up the sign of the index with the sign of the value. How to avoid it: a^-n = 1/a^n (reciprocal), not -a^n. E.g. 2^-3 = 1/8, not -8.

Why is "Cancelling individual terms in a fraction (e.g. $\dfrac{x + 2}{x + 3} = \dfrac{2}{3}$)" a common mistake in Algebraic Expressions?

Why it happens: Treating a sum like a product. How to avoid it: You may only cancel COMMON FACTORS, not common terms. x + 2/x + 3 does NOT simplify. 2(x + 2)/3(x + 2) = 2/3 DOES — because (x + 2) is a factor of both numerator and denominator. Source: WMA11/01 examiner reports — flagged annually

Why is "Saying $(x + 2)$ is a factor when $f(2) = 0$" a common mistake in Algebraic Expressions?

Why it happens: Sign confusion in the factor theorem. How to avoid it: (x - a) is a factor f(a) = 0. So f(2) = 0 means (x - 2) is a factor. f(-2) = 0 means (x + 2) is a factor.

Pure Mathematics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Algebraic Expressions

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Algebraic Expressions Study Notes — Edexcel IAL Mathematics P1 (WMA11/01, 2018 spec — 2026 onwards)

Indices, surds, expansion, factorisation and algebraic fractions. The IAL P1 foundation: every later topic — calculus, trig, vectors — leans on fluent algebra. Calculator allowed but expect exact-form (surd) answers.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

1.1 — Understand and use the laws of indices for all rational exponents.
1.2 — Use and manipulate surds, including rationalising the denominator.
1.3 — Manipulate polynomials algebraically, including expanding brackets and collecting like terms, factorisation, simple algebraic division.
1.4 — Simplify rational expressions including by factorising and cancelling, and algebraic division (by linear expressions only).
1.5 — Use the factor theorem and the algebraic relationship between coefficients and roots.

Laws of indices — including fractional and negative

Six rules. Drill them until they're automatic.

The six laws of indices (none of these are in the IAL formula booklet — you must memorise):

Rule	Statement	Example
Product	$a^{m} \cdot a^{n} = a^{m+n}$	$x^{3} \cdot x^{5} = x^{8}$
Quotient	$a^{m} / a^{n} = a^{m-n}$	$x^{7} / x^{2} = x^{5}$
Power of a power	$(a^{m})^{n} = a^{mn}$	$(x^{2})^{3} = x^{6}$
Zero index	$a^{0} = 1$ (for $a \neq 0$ )	$5^{0} = 1$
Negative index	$a^{-n} = 1/a^{n}$	$x^{-2} = 1/x^{2}$
Fractional index	$a^{m/n} = \sqrt[n]{a^{m}}$	$8^{2/3} = (\sqrt[3]{8})^{2} = 4$

Fractional indices. $a^{p/q}$ means: take the $q$ -th root, then raise to the $p$ -th power. Equivalent to raise-then-root, but root-first usually keeps the numbers small.

$27^{4/3} = (\sqrt[3]{27})^{4} = 3^{4} = 81$

Negative indices = reciprocals, not negative values.

$2^{-3} = \dfrac{1}{2^{3}} = \dfrac{1}{8} \text{ (NOT } -8 \text{)}$

Combining rules. $\left(\dfrac{16 x^{8}}{y^{4}}\right)^{-3/4}$ :

Negative index flips: $\left(\dfrac{y^{4}}{16 x^{8}}\right)^{3/4}$ .
Distribute power: $\dfrac{y^{3}}{16^{3/4} x^{6}}$ .
Compute $16^{3/4} = (16^{1/4})^{3} = 2^{3} = 8$ .
Answer: $\dfrac{y^{3}}{8 x^{6}}$ .

Product: ADD indices.
Quotient: SUBTRACT indices.
Power of power: MULTIPLY indices.
Negative: FLIP fraction.
Fractional: ROOT first, then power.

Surds — simplifying, multiplying, rationalising

Pull out largest square factor. Conjugate kills $a + b\sqrt{c}$ .

A surd is an irrational root that can't be reduced to a rational number. $\sqrt{2}$ , $\sqrt{15}$ , $\sqrt[3]{4}$ . But $\sqrt{9} = 3$ is NOT a surd.

Simplifying. Pull out the largest perfect-square factor.

$\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \qquad \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}.$

Rules (NOT in formula booklet):

$\sqrt{ab} = \sqrt{a}\sqrt{b}$
$\sqrt{a/b} = \sqrt{a}/\sqrt{b}$
$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$ (this is wrong — see pitfall)

Rationalising — single surd in denominator. Multiply top and bottom by the same surd:

$\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}.$

Rationalising — binomial denominator. Multiply by the conjugate: switch the sign between the rational and surd part.

$\dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \dfrac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$

The trick: $(a + b\sqrt{c})(a - b\sqrt{c}) = a^{2} - b^{2}c$ is rational.

Multiplying by the conjugate of 2 + √3 gives a rational denominator using a² − b²c.

Worked example. Rationalise $\dfrac{3 + \sqrt{5}}{2 - \sqrt{5}}$ .

Multiply by $(2 + \sqrt{5})/(2 + \sqrt{5})$ .
Numerator: $(3 + \sqrt{5})(2 + \sqrt{5}) = 6 + 3\sqrt{5} + 2\sqrt{5} + 5 = 11 + 5\sqrt{5}$ .
Denominator: $(2)^{2} - (\sqrt{5})^{2} = 4 - 5 = -1$ .
Answer: $\dfrac{11 + 5\sqrt{5}}{-1} = -11 - 5\sqrt{5}$ .

Edexcel tip. Exam questions often say "in the form $a + b\sqrt{c}$ where $a$ , $b$ , $c$ are integers" — that's a strong hint to rationalise.

Largest square factor → simplify.
Single surd denominator → multiply by that surd.
Binomial denominator → multiply by conjugate.
$(a + b\sqrt{c})(a - b\sqrt{c}) = a^{2} - b^{2}c$ (rational).

Expanding brackets — single, double, triple

Distributive law. Stay systematic. Always collect like terms.

Single bracket. Distribute.

$3x(2x - 5) = 6x^{2} - 15x.$

Double bracket (FOIL).

$(x + 4)(x - 3) = x^{2} - 3x + 4x - 12 = x^{2} + x - 12.$

Triple bracket. Expand the first two, then multiply by the third.

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

Step 1: $(x - 2)(x + 3) = x^{2} + x - 6$ .

Step 2: $(x^{2} + x - 6)(2x - 1) = 2x^{3} - x^{2} + 2x^{2} - x - 12x + 6 = 2x^{3} + x^{2} - 13x + 6$ .

Mark-scheme awarding (typical 3-mark question).

Mark	Awarded for
M1	Correct expansion of any two brackets
M1	Multiplying the result by the third bracket (ECF allowed)
A1	Correct, fully simplified answer

Binomial expansion shortcut. For $(a + b)^{2}$ and $(a + b)^{3}$ , use:

$(a + b)^{2} = a^{2} + 2ab + b^{2}$
$(a - b)^{2} = a^{2} - 2ab + b^{2}$
$(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$
$(a - b)^{3} = a^{3} - 3a^{2}b + 3ab^{2} - b^{3}$

Only the first two appear regularly in P1; the cube forms are useful for P2's general binomial theorem.

Distribute carefully — track signs.
Two brackets first, then third.
Collect like terms last.
Memorise $(a \pm b)^{2}$ .

Factor theorem — factorising cubics

Find one root by trial. Divide. Factorise the quadratic remainder.

Factor theorem. $(x - a)$ is a factor of $f(x)$ $\Leftrightarrow$ $f(a) = 0$ .

Strategy for factorising a cubic $f(x) = ax^{3} + bx^{2} + cx + d$ :

Try integer factors of $d$ (the constant term) — $\pm 1, \pm 2, \pm 3, \ldots$ . Most exam questions have one obvious root.
Divide $f(x)$ by $(x - a)$ to get a quadratic.
Factorise the quadratic.

Worked example. $f(x) = 2x^{3} + 3x^{2} - 8x + 3$ .

Try $x = 1$ : $f(1) = 2 + 3 - 8 + 3 = 0$ . ✓ So $(x - 1)$ is a factor.
Divide: $f(x) = (x - 1)(2x^{2} + 5x - 3)$ .
Factorise the quadratic: $2x^{2} + 5x - 3 = (2x - 1)(x + 3)$ .
Final: $f(x) = (x - 1)(2x - 1)(x + 3)$ .

Polynomial division. Either long division, or comparing coefficients:

If $2x^{3} + 3x^{2} - 8x + 3 = (x - 1)(2x^{2} + Bx + C)$ , expand the right: $2x^{3} + Bx^{2} + Cx - 2x^{2} - Bx - C = 2x^{3} + (B - 2)x^{2} + (C - B)x - C.$

Match coefficients:

$x^{2}$ : $B - 2 = 3 \Rightarrow B = 5$ .
$x^{0}$ : $-C = 3 \Rightarrow C = -3$ .
Check $x^{1}$ : $C - B = -3 - 5 = -8$ . ✓

Test factors of constant term: $\pm 1, \pm 2, \ldots$ .
$f(a) = 0 \Rightarrow (x - a)$ is a factor.
Divide or compare coefficients to get the quadratic.
Factorise the quadratic to get full factorisation.

Algebraic fractions — factorise, cancel, state restrictions

Cancel COMMON FACTORS, never individual terms.

Rule of cancelling. You may only cancel common factors (things multiplied), never common terms (things added).

Allowed	Not allowed
$\dfrac{2(x + 3)}{5(x + 3)} = \dfrac{2}{5}$	$\dfrac{2 + (x + 3)}{5 + (x + 3)} = \dfrac{2}{5}$ ❌
$\dfrac{x(x - 1)}{x + 2}$ — no cancel	$\dfrac{x + 1}{x + 2} = \dfrac{1}{2}$ ❌

Method for simplifying:

Fully factorise the numerator.
Fully factorise the denominator.
Cancel any factor appearing in both.

Worked example. Simplify $\dfrac{2x^{2} + 5x - 3}{x^{2} - 9}$ .

Numerator: $2x^{2} + 5x - 3 = (2x - 1)(x + 3)$ .
Denominator: $x^{2} - 9 = (x - 3)(x + 3)$ .
Cancel $(x + 3)$ : $\dfrac{2x - 1}{x - 3}$ for $x \neq -3$ .

Adding/subtracting algebraic fractions. Common denominator first.

$\dfrac{2}{x + 1} + \dfrac{3}{x - 2} = \dfrac{2(x - 2) + 3(x + 1)}{(x + 1)(x - 2)} = \dfrac{5x - 1}{(x + 1)(x - 2)}.$

Multiplying. Multiply numerators and denominators, factorise everything, cancel.

$\dfrac{x + 2}{x - 1} \cdot \dfrac{x^{2} - 1}{x + 2} = \dfrac{(x + 2)(x - 1)(x + 1)}{(x - 1)(x + 2)} = x + 1.$

Dividing. Flip and multiply.

$\dfrac{x + 2}{x - 1} \div \dfrac{x + 2}{x + 1} = \dfrac{x + 2}{x - 1} \cdot \dfrac{x + 1}{x + 2} = \dfrac{x + 1}{x - 1}.$

Factorise everything first.
Cancel common FACTORS only.
Adding/subtracting: common denominator.
Dividing: flip and multiply.

How it’s examined

Algebraic Expressions appears in WMA11/01 P1 every sitting (Jan & Jun), usually as Q1-Q5 (warm-up marks). Typical questions: simplify a surd or rationalise (3-4 marks); simplify an index expression (3 marks); use the factor theorem to factorise a cubic (5-6 marks); simplify an algebraic fraction (3-4 marks). Mark schemes use M (method), A (accuracy), B (independent) awarding codes. ECF (error carried forward) applies generously — get the structure right and you keep most marks even with arithmetic slips. A* students should aim for full marks on this section — it is the easiest part of P1.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA11/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Algebraic Expressions

Step-by-step solutions to past-paper-style questions on algebraic expressions, written exactly the way a tutor would explain them at the board.

1Rationalise the denominator (4 marks, P1)
Core• Style: WMA11/01 January 2024 Q1• surds, rationalising
Question
Express $\dfrac{3 + \sqrt{5}}{2 - \sqrt{5}}$ in the form $a + b\sqrt{5}$ , where $a$ and $b$ are integers.
Step-by-step solution
1. Step 1
  Multiply top and bottom by the conjugate of the denominator, $(2 + \sqrt{5})$ .
  $\dfrac{3 + \sqrt{5}}{2 - \sqrt{5}} \times \dfrac{2 + \sqrt{5}}{2 + \sqrt{5}}$
2. Step 2
  Expand the numerator: $(3 + \sqrt{5})(2 + \sqrt{5}) = 6 + 3\sqrt{5} + 2\sqrt{5} + 5 = 11 + 5\sqrt{5}$ .
3. Step 3
  Expand the denominator (difference of two squares): $(2 - \sqrt{5})(2 + \sqrt{5}) = 4 - 5 = -1$ .
4. Step 4
  Divide:
  $\dfrac{11 + 5\sqrt{5}}{-1} = -11 - 5\sqrt{5}$
Answer
$a = -11$ , $b = -5$ , i.e. $-11 - 5\sqrt{5}$ .
Examiner tip
Mark scheme: M1 for multiplying by the correct conjugate, M1 for an attempt at expansion (either numerator OR denominator), A1 for the numerator simplified, A1 for the final answer. A common ECF: students forget the minus in $4 - 5 = -1$ and write $1$ , then lose only the final A1.
2Fractional and negative indices (3 marks, P1)
Core• Style: WMA11/01 June 2024 Q2• indices
Question
Simplify $\left(\dfrac{16 x^{8}}{y^{4}}\right)^{-\tfrac{3}{4}}$ , giving your answer in the form $\dfrac{y^{p}}{k x^{q}}$ where $k$ , $p$ , $q$ are positive integers.
Step-by-step solution
1. Step 1
  Negative power flips the fraction:
  $\left(\dfrac{16 x^{8}}{y^{4}}\right)^{-3/4} = \left(\dfrac{y^{4}}{16 x^{8}}\right)^{3/4}$
2. Step 2
  Apply the power to each factor:
  $\dfrac{(y^{4})^{3/4}}{(16)^{3/4}\,(x^{8})^{3/4}}$
3. Step 3
  Compute each: $(y^{4})^{3/4} = y^{3}$ ; $(x^{8})^{3/4} = x^{6}$ ; $16^{3/4} = (16^{1/4})^{3} = 2^{3} = 8$ .
4. Step 4
  Assemble:
  $\dfrac{y^{3}}{8 x^{6}}$
Answer
$\dfrac{y^{3}}{8 x^{6}}$ (so $k = 8$ , $p = 3$ , $q = 6$ ).
Examiner tip
M1 for handling the negative sign (flipping the fraction OR writing as reciprocal). M1 for splitting power across factors. A1 for $16^{3/4} = 8$ . A1 for fully simplified answer. The most common dropped mark is computing $16^{3/4}$ : students who write $16^{3/4} = 12$ lose A1 immediately.
3Expand three brackets (3 marks, P1)
Core• Style: WMA11/01 January 2023 Q3• expansion
Question
Expand and simplify $(x - 2)(x + 3)(2x - 1)$ .
Step-by-step solution
1. Step 1
  Expand the first two brackets:
  $(x - 2)(x + 3) = x^{2} + x - 6$
2. Step 2
  Now multiply by $(2x - 1)$ :
  $(x^{2} + x - 6)(2x - 1) = 2x^{3} - x^{2} + 2x^{2} - x - 12x + 6$
3. Step 3
  Collect like terms: $2x^{3} + (-1 + 2)x^{2} + (-1 - 12)x + 6 = 2x^{3} + x^{2} - 13x + 6$ .
Answer
$2x^{3} + x^{2} - 13x + 6$
Examiner tip
M1 for any correct expansion of two brackets. M1 for multiplying the result by the third. A1 for the fully correct expansion. ECF applies: if you slip on the first product but multiply your wrong quadratic correctly by $(2x - 1)$ , you still earn the second M.
4Factorise a cubic using the factor theorem (5 marks, P1)
Extended• Style: WMA11/01 June 2023 Q5• factor theorem, polynomial division
Question
$f(x) = 2x^{3} + 3x^{2} - 8x + 3$ . Show that $(x - 1)$ is a factor of $f(x)$ , and hence factorise $f(x)$ fully.
Step-by-step solution
1. Step 1
  Evaluate $f(1)$ :
  $f(1) = 2 + 3 - 8 + 3 = 0$
2. Step 2
  Since $f(1) = 0$ , by the factor theorem $(x - 1)$ is a factor.
3. Step 3
  Divide $f(x)$ by $(x - 1)$ (long division or compare coefficients). The quotient is a quadratic $2x^{2} + ax + b$ . Comparing $x^{0}$ : $b \cdot (-1) = 3$ so $b = -3$ . Comparing $x^{0}$ in product: $(2x^{2} + 5x - 3)(x - 1) = 2x^{3} - 2x^{2} + 5x^{2} - 5x - 3x + 3 = 2x^{3} + 3x^{2} - 8x + 3$ . ✓
4. Step 4
  So $f(x) = (x - 1)(2x^{2} + 5x - 3)$ .
5. Step 5
  Factorise the quadratic: $2x^{2} + 5x - 3 = (2x - 1)(x + 3)$ .
6. Step 6
  Full factorisation:
  $f(x) = (x - 1)(2x - 1)(x + 3)$
Answer
$f(x) = (x - 1)(2x - 1)(x + 3)$
Examiner tip
B1 for evaluating $f(1) = 0$ with conclusion. M1 for attempting division/comparison. A1 for the quadratic factor. M1 for attempting to factorise the quadratic. A1 for the full factorisation. NB: 'Show that' demands that you actually substitute $x = 1$ and state the factor-theorem conclusion in words — just writing ' $f(1) = 0$ ' without naming the theorem can cost B1.
5Simplify an algebraic fraction (4 marks, P1)
Extended• Style: WMA11/01 January 2024 Q4• algebraic fractions
Question
Simplify fully $\dfrac{2x^{2} + 5x - 3}{x^{2} - 9}$ .
Step-by-step solution
1. Step 1
  Factorise the numerator. Looking for two numbers with product $2 \times (-3) = -6$ and sum $5$ : $6$ and $-1$ . Split: $2x^{2} + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)$ .
2. Step 2
  Factorise the denominator as a difference of two squares: $x^{2} - 9 = (x - 3)(x + 3)$ .
3. Step 3
  Cancel the common factor $(x + 3)$ :
  $\dfrac{(2x - 1)(x + 3)}{(x - 3)(x + 3)} = \dfrac{2x - 1}{x - 3}$
Answer
$\dfrac{2x - 1}{x - 3}$ (provided $x \neq -3$ )
Examiner tip
M1 for factorising the numerator (correct quadratic factors). M1 for factorising the denominator. A1 for cancelling the common factor. A1 for the simplified fraction in lowest terms. Strictly the answer requires $x \neq -3$ ; mark schemes don't always demand this, but quoting it shows mathematical maturity.

Key Formulae — Algebraic Expressions

The formulae you need to memorise for algebraic expressions on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Laws of indices
$a^{m} \cdot a^{n} = a^{m+n}, \quad \dfrac{a^{m}}{a^{n}} = a^{m-n}, \quad (a^{m})^{n} = a^{mn}, \quad a^{-n} = \dfrac{1}{a^{n}}, \quad a^{m/n} = \sqrt[n]{a^{m}}, \quad a^{0} = 1$
$a$
base (any non-zero real number)
$m, n$
indices (any rational numbers)
When to use
Whenever you simplify expressions with powers. NOT in the IAL formula booklet — these must be memorised.
Example
$x^{1/2} \cdot x^{1/3} = x^{5/6}$ .
Surd rules
$\sqrt{ab} = \sqrt{a}\sqrt{b}, \quad \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}, \quad \dfrac{1}{\sqrt{a}} = \dfrac{\sqrt{a}}{a}, \quad \dfrac{1}{a + \sqrt{b}} = \dfrac{a - \sqrt{b}}{a^{2} - b}$
$a, b$
non-negative reals (with $b > 0$ )
When to use
When simplifying or rationalising. NOT in the formula booklet.
Example
$\dfrac{6}{\sqrt{12}} = \dfrac{6}{2\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$ .
Difference of two squares
$a^{2} - b^{2} = (a - b)(a + b)$
$a, b$
any expressions
When to use
Factorising and rationalising. NOT in the formula booklet.
Example
$x^{2} - 25 = (x - 5)(x + 5)$ ; also used for surd conjugates.
Factor theorem
$f(a) = 0 \Leftrightarrow (x - a) \text{ is a factor of } f(x)$
$f(x)$
polynomial
$a$
a real number
When to use
To factorise a cubic when one root can be spotted (try $\pm 1, \pm 2,$ factors of constant term). NOT in the formula booklet.
Example
If $f(2) = 0$ , then $(x - 2)$ is a factor of $f(x)$ .

Key Definitions and Keywords — Algebraic Expressions

Definitions to memorise and the exact keywords mark schemes credit for algebraic expressions answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Surd
Examiner keyword
An irrational root that cannot be simplified to a rational number (e.g. $\sqrt{2}$ , $\sqrt{15}$ ). $\sqrt{9} = 3$ is NOT a surd.
Example
$\sqrt{2}$ , $3\sqrt{5}$ , $\sqrt{7} + 1$ .
Rationalise the denominator
Examiner keyword
Rewrite a fraction so its denominator contains no surds. Multiply top and bottom by the appropriate surd or conjugate.
Example
$\dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \dfrac{2 - \sqrt{3}}{1}$ .
Conjugate (of a surd binomial)
The conjugate of $a + b\sqrt{c}$ is $a - b\sqrt{c}$ . Their product is rational: $a^{2} - b^{2}c$ .
Example
Conjugate of $3 + \sqrt{2}$ is $3 - \sqrt{2}$ ; their product is $9 - 2 = 7$ .
Polynomial
An expression of the form $a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{0}$ where all $a_{i}$ are real constants and $n$ is a non-negative integer.
Example
$2x^{3} + 5x - 7$ is a polynomial of degree 3.
Factor
$(x - a)$ is a factor of $f(x)$ if $f(x)$ can be written as $(x - a) \cdot g(x)$ for some polynomial $g(x)$ . Equivalent to $f(a) = 0$ .
Related: factor theorem

Common Mistakes and Misconceptions — Algebraic Expressions

The traps other students keep falling into on algebraic expressions questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Writing $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$
WMA11/01 examiner reports flag this regularly
Why it happens
Over-extending the product rule $\sqrt{ab} = \sqrt{a}\sqrt{b}$ to sums.
How to avoid it
Test with numbers: $\sqrt{9 + 16} = \sqrt{25} = 5$ , NOT $3 + 4 = 7$ . The square root does NOT distribute over addition. Only over multiplication and division.
✕Computing $16^{3/4}$ as $16 \div (3/4) = 21.33$ or as $12$
Why it happens
Confusing index notation with division.
How to avoid it
$a^{m/n}$ means 'take the $n$ -th root, then raise to the $m$ -th power'. So $16^{3/4} = (16^{1/4})^{3} = 2^{3} = 8$ . Always take the root FIRST.
✕Treating $a^{-n}$ as $-a^{n}$
Why it happens
Mixing up the sign of the index with the sign of the value.
How to avoid it
$a^{-n} = \dfrac{1}{a^{n}}$ (reciprocal), not $-a^{n}$ . E.g. $2^{-3} = \dfrac{1}{8}$ , not $-8$ .
✕Cancelling individual terms in a fraction (e.g. $\dfrac{x + 2}{x + 3} = \dfrac{2}{3}$ )
WMA11/01 examiner reports — flagged annually
Why it happens
Treating a sum like a product.
How to avoid it
You may only cancel COMMON FACTORS, not common terms. $\dfrac{x + 2}{x + 3}$ does NOT simplify. $\dfrac{2(x + 2)}{3(x + 2)} = \dfrac{2}{3}$ DOES — because $(x + 2)$ is a factor of both numerator and denominator.
✕Saying $(x + 2)$ is a factor when $f(2) = 0$
Why it happens
Sign confusion in the factor theorem.
How to avoid it
$(x - a)$ is a factor $\Leftrightarrow f(a) = 0$ . So $f(2) = 0$ means $(x - 2)$ is a factor. $f(-2) = 0$ means $(x + 2)$ is a factor.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Algebraic Expressions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Algebraic Expressions

Short Study Notes in the form of Slides

Short Study Notes — Algebraic Expressions

Detailed Notes

What you’ll learn

How it’s examined

1Rationalise the denominator (4 marks, P1)

2Fractional and negative indices (3 marks, P1)

3Expand three brackets (3 marks, P1)

4Factorise a cubic using the factor theorem (5 marks, P1)

5Simplify an algebraic fraction (4 marks, P1)

Laws of indices

Surd rules

Difference of two squares

Factor theorem

Surd

Rationalise the denominator

Conjugate (of a surd binomial)

Polynomial

Factor

✕Writing a+b=a+b\sqrt{a + b} = \sqrt{a} + \sqrt{b}a+b​=a​+b​

✕Computing 163/416^{3/4}163/4 as 16÷(3/4)=21.3316 \div (3/4) = 21.3316÷(3/4)=21.33 or as 121212

✕Treating a−na^{-n}a−n as −an-a^{n}−an

✕Cancelling individual terms in a fraction (e.g. x+2x+3=23\dfrac{x + 2}{x + 3} = \dfrac{2}{3}x+3x+2​=32​)

✕Saying (x+2)(x + 2)(x+2) is a factor when f(2)=0f(2) = 0f(2)=0

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Algebraic Expressions — frequently asked questions

✕Writing $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$

✕Computing $16^{3/4}$ as $16 \div (3/4) = 21.33$ or as $12$

✕Treating $a^{-n}$ as $-a^{n}$

✕Cancelling individual terms in a fraction (e.g. $\dfrac{x + 2}{x + 3} = \dfrac{2}{3}$ )

✕Saying $(x + 2)$ is a factor when $f(2) = 0$