A Algebra · notation and manipulation

A1

Use and interpret algebraic manipulation: ab for a × b, 3y for y + y + y and 3 × y, a² for a × a, a³ for a × a × a, a²b for a × a × b, a/b for a ÷ b, coefficients as fractions, brackets

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Algebraic notation records operations compactly. Adjacent letters mean multiplication, so $ab=a\times b$ , while $3y$ means three lots of $y$ .
An index records repeated factors: $a^2=a\times a$ and $a^2b=a\times a\times b$ .
A fraction bar represents division and also groups its numerator and denominator.
Write numerical coefficients before variables, usually as exact fractions rather than decimals, and use brackets when an operation acts on a complete expression.
Examiners expect conventional notation and the operations to remain unambiguous when translating words or repeated products.

Worked example

A rectangle has length $\dfrac{3x}{2}$ and width $y$ . Write its area and perimeter in conventional algebraic notation.

1.Area $=\dfrac{3x}{2}\times y=\dfrac{3xy}{2}$ .
2.Perimeter $=2\left(\dfrac{3x}{2}\right)+2y$ .
3.Simplify to $3x+2y$ .

Answer: Area $=\dfrac{3xy}{2}$ and perimeter $=3x+2y$ .

Common mistakes

Don't read $a^2$ as $2a$ instead of $a\times a$ .
Don't interpret $3y$ as $3+y$ rather than $3\times y$ .
Don't drop brackets when a multiplier must act on a whole expression.

Exam tip

Translate one operation at a time and use brackets before simplifying the notation.

Tier 1 · Easy

ORIGINAL

1

Write

m\times m\times n\times n\times n

using indices.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Write

p\times p\times p\times q\div5

in conventional algebraic notation, and state its coefficient.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A rectangle has length

\frac{3x}{2}

and width

y

. Write its area and its perimeter in conventional algebraic notation.

(3)

(Total for Question 1 is 3 marks)

A2

Substitute numerical values into formulae and expressions, including scientific formulae

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Substitution replaces every occurrence of a variable with its given value while preserving the original operations. Put negative and fractional values in brackets so that powers and signs act on the whole value.
Follow the order of operations: evaluate powers before multiplication, division, addition and subtraction.
In a scientific formula, include the stated units and convert them first if necessary.
Keep full calculator precision until the requested rounding.
Examiners award method for a correct substituted expression, so write that line before evaluating rather than giving only a calculator answer.

Worked example

The kinetic energy formula is $E=\dfrac12mv^2$ . Find $E$ when $m=3.2\,\text{kg}$ and $v=5\,\text{m s}^{-1}$ .

1.Substitute both values: $E=\dfrac12\times3.2\times5^2$ .
2.Evaluate the power first: $5^2=25$ .
3. $E=0.5\times3.2\times25=40\,\text{J}$ .

Answer: $40\,\text{J}$ .

Common mistakes

Don't substitute $v=5$ into $v^2$ as $2\times5$ .
Don't write $(-3)^2$ as $-9$ after omitting the brackets.
Don't round an intermediate value and lose accuracy in the final answer.

Exam tip

Show the formula with every value substituted before entering it into the calculator.

Tier 1 · Easy

ORIGINAL

1

Work out

2x^2-5

when

x=-3

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

The kinetic energy of an object is given by

E=\frac12mv^2

. Work out

E

when

m=3.2\,\text{kg}

and

v=5\,\text{m s}^{-1}

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Use

E=mc^2

to calculate

E

when

m=4.2\times10^{-8}\,\text{kg}

and

c=3\times10^8\,\text{m s}^{-1}

. Give your answer in standard form.

(3)

(Total for Question 1 is 3 marks)

A3

Understand and use the concepts and vocabulary of expressions, equations, formulae, identities, inequalities, terms and factors

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

An expression has no equality or inequality sign. An equation is true only for particular values, whereas an identity is true for every permitted value and is written with $\equiv$ .
A formula links quantities, and an inequality compares a range of possible values.
Terms are separated by addition or subtraction; factors are quantities multiplied together.
For example, $6x^2-15x$ has two terms and factorises as $3x(2x-5)$ , whose factors are $3x$ and $2x-5$ .
Examiners expect the correct vocabulary and a reason based on the statement's structure or truth.

Worked example

Classify $3(2x-1)=9$ , $3(2x-1)\equiv6x-3$ , and $3(2x-1)<9$ .

1. $3(2x-1)=9$ is true only for a particular value, so it is an equation.
2.Expanding $3(2x-1)$ always gives $6x-3$ , so the second statement is an identity.
3.The symbol $<$ compares possible values, so the third statement is an inequality.

Answer: Equation, identity, inequality, in that order.

Common mistakes

Don't call every statement containing an equals sign an identity.
Don't count factors as terms even though terms are separated by addition or subtraction.
Don't use $=$ instead of $\equiv$ for a relationship true for all permitted values.

Exam tip

When asked to classify a statement, justify whether it is always true, sometimes true or a comparison.

Tier 1 · Easy

ORIGINAL

1

State whether

5x+7

is an expression, equation or inequality.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

For

8x^2-12x=4x(2x-3)

, state the number of terms on the left and name the two factors on the right.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Classify each statement as an equation, an identity or an inequality:

3(2x-1)=9

,

3(2x-1)\equiv6x-3

, and

3(2x-1)<9

.

(3)

(Total for Question 1 is 3 marks)

A4

Simplify and manipulate algebraic expressions (incl. surds and algebraic fractions): like terms, common factors, expanding two or more binomials, factorising quadratics incl. ax² + bx + c, indices

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Specification A4: simplify and manipulate algebraic expressions, including those involving surds and algebraic fractions, by collecting like terms, multiplying a single term over a bracket, taking out common factors, expanding products of two or more binomials, factorising quadratic expressions of the form $x^2+bx+c$ , including the difference of two squares, and, at Higher tier, $ax^2+bx+c$ , and simplifying expressions involving sums, products and powers, including the laws of indices.

Explanation

Collect only like terms and apply index laws only to matching bases: $x^3\times x^5=x^8$ , $(x^3)^5=x^{15}$ . To expand two binomials, multiply every term in the first bracket by every term in the second, giving four terms, then collect; the method mark is usually for at least three of the four terms correct.
To factorise, first take out the highest common factor of every term, including any letters, then check by expanding. For $x^2+bx+c$ find two numbers with product $c$ and sum $b$ .
Higher tier: for $ax^2+bx+c$ either split the middle term or test bracket pairs whose outer and inner products sum to $bx$ ; a common factor removed first makes the remaining quadratic easier.
Never cancel across an addition in an algebraic fraction; factorise the numerator and denominator completely and cancel whole factors.
A fully factorised answer has no common factor left inside any bracket.

Another way to see this:StandardVisual

Worked example

1

Factorise fully

6x^2+7x-3

.

(2)

(Total for Question 1 is 2 marks)

Working

1.The product of the outer coefficients is $6\times(-3)=-18$ and the middle coefficient is $7$ ; the pair $9$ and $-2$ has product $-18$ and sum $7$ .
2.Split the middle term: $6x^2+9x-2x-3$ .
3.Factorise in pairs: $3x(2x+3)-1(2x+3)=(2x+3)(3x-1)$ .
4.Check by expanding: $6x^2-2x+9x-3=6x^2+7x-3$ .

Answer: $(2x+3)(3x-1)$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for $(2x\pm3)(3x\pm1)$ or $(2x\pm1)(3x\pm3)$ or $(6x\pm3)(x\pm1)$ or $(6x\pm1)(x\pm3)$	M1	or for splitting the middle term correctly, eg $6x^2+9x-2x-3$
	for $(2x+3)(3x-1)$	A1	cao

Common mistakes

Don't collect unlike terms such as $3x+2x^2$ to make $5x^3$ .
Don't miss a cross-term when expanding two brackets.
Don't cancel terms across addition in an algebraic fraction.

Mark-scheme wording to use

“M1 for at least 3 terms out of a maximum of 4 correct from expansion” — Pearson 1MA1, Sample Paper 1H, Q2
“M1 for 3(5x^3 + x^2y) or x(15x^2+3xy) or 3x(5x^2 + xy) or x^2(15x + 3y) or 3x^2(ax + by) Where a ≥ 1 and b ≥ 1 A1 cao” — Pearson 1MA1, June 2022 Paper 2H, Q1(c)
“M1 (x ± 1)(x ± 4) A1 (x − 1)(x + 4) oe” — Pearson 1MA1, Specimen 1 Paper 3H, Q6

What the examiner wants

A finished expression. The June 2024 Paper 3H report records that factorising only the first two terms and giving $k(3k+11)-4$ was a popular error that scored nothing, and the November 2024 Paper 3H report that a final answer left as a sum, or with multiplication signs still in it, could not gain full marks.

Exam tip

After factorising, expand your answer mentally; it must reproduce the original expression exactly.

Tier 1 · Easy

ORIGINAL

1

Simplify

7a+3b-2a+5b

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Factorise

6x^2+x-2

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Simplify

\frac{x^2-9}{x^2+x-6}

, stating every value of

x

excluded from the original expression.

(4)

(Total for Question 1 is 4 marks)

A5

Understand and use standard mathematical formulae; rearrange formulae to change the subject

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

The subject of a formula is the variable isolated on one side. Changing the subject must preserve an equivalent relationship, so perform the same operation on both sides and undo operations in reverse order.
Clear fractions or brackets when this makes the structure easier to see.
Foundation questions usually isolate a subject that appears once.
Higher tier: the subject may appear more than once or inside a fraction, requiring expansion and collection of its terms.
Examiners expect each inverse operation to be visible and the final subject to appear alone.

Worked example

Make $t$ the subject of $v=u+at$ .

1.Subtract $u$ from both sides: $v-u=at$ .
2.Divide both sides by $a$ : $\dfrac{v-u}{a}=t$ .
3.Write the subject first: $t=\dfrac{v-u}{a}$ .

Answer: $t=\dfrac{v-u}{a}$ .

Common mistakes

Don't change a sign while moving a term without applying an operation to both sides.
Don't divide only one term of a sum instead of the entire side.
Don't stop with the requested subject still multiplied by another quantity.

Exam tip

State one balancing operation per line until the requested subject is alone.

Tier 1 · Easy

ORIGINAL

1

Make

w

the subject of

A=lw

.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

The area of a trapezium is given by

A=\dfrac12(a+b)h

, where

A

is measured in

\text{cm}^2

and

a

,

b

and

h

are measured in cm. (a) Make

h

the subject of the formula. (b) Hence work out

h

when

A=45\,\text{cm}^2

,

a=7\,\text{cm}

and

b=11\,\text{cm}

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Make

a

the subject of

P=\frac{a+b}{a-b}

.

(4)

(Total for Question 1 is 4 marks)

Use and interpret algebraic manipulation: ab for a × b, 3y for y + y + y and 3 × y, a² for a × a, a³ for a × a × a, a²b for a × a × b, a/b for a ÷ b, coefficients as fractions, brackets

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Substitute numerical values into formulae and expressions, including scientific formulae

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Understand and use the concepts and vocabulary of expressions, equations, formulae, identities, inequalities, terms and factors

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Simplify and manipulate algebraic expressions (incl. surds and algebraic fractions): like terms, common factors, expanding two or more binomials, factorising quadratics incl. ax² + bx + c, indices

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Understand and use standard mathematical formulae; rearrange formulae to change the subject

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Want help turning these notes into marks?