Expanding Brackets

Topic

Higher June 2025 Paper 1R Q17

EdexcelCurrent spec3 marksExpanding Brackets

17 Prove that, for any three numbers which are consecutive multiples of 4, the difference between the square of the largest number and the square of the smallest number is always a multiple of 64

Show clear algebraic working.

(3)

Mark scheme

Scheme	Marks
eg \(4n, 4n + 4, 4n + 8\) or \(4n, 4(n + 1), 4(n + 2)\) or \(4n - 4, 4n, 4n + 4\) or \(4(n - 1), 4n, 4(n + 1)\)	M1
eg \((4n + 8)^2 - (4n)^2\) (= \(16n^2 + 64n + 64 - 16n^2\)) or \((4(n + 2))^2 - (4n)^2\) (= \(16n^2 + 64n + 64 - 16n^2\)) or \((4n + 4)^2 - (4n - 4)^2\) (= \(16n^2 + 32n + 16 - 16n^2 + 32n - 16\)) or \((4(n + 1))^2 - (4(n - 1))^2\) (= \(16n^2 + 32n + 16 - 16n^2 + 32n - 16\))	M1
eg \((4n + 8)^2 - (4n)^2 = 16n^2 + 64n + 64 - 16n^2 = 64n + 64\) or \((4n + 8)^2 - (4n)^2 = (4n + 8 + 4n)(4n + 8 - 4n) = 8(8n + 8) = 64n + 64\) or \((4(n + 2))^2 - (4n)^2 = 16n^2 + 64n + 64 - 16n^2 = 64n + 64\) or \((4(n + 2))^2 - (4n)^2 = (4(n + 2) + 4n)(4(n + 2) - 4n) = 8(8n + 8) = 64n + 64\) or \((4n + 4)^2 - (4n - 4)^2 = 16n^2 + 32n + 16 - 16n^2 + 32n - 16 = 64n\) or \((4n + 4)^2 - (4n - 4)^2 = (4n + 4 + 4n - 4)(4n + 4 - 4n + 4) = 8n \times 8 = 64n\) or \((4(n + 1))^2 - (4(n - 1))^2 = 16n^2 + 32n + 16 - 16n^2 + 32n - 16 = 64n\) or \((4(n + 1))^2 - (4(n - 1))^2 = (4(n + 1) + 4(n - 1))(4(n + 1) - 4(n - 1)) = 8n \times 8 = 64n\) Working required Answer: correctly shown	A1
	(3)
	(3 marks)

Notes

M1: for correct expressions for 3 consecutive multiples of 4 (any letter can be used) may just see the first and third multiple for this mark

M1: for squaring the largest and smallest multiple of 4 and subtracting (no need to expand or simplify for this mark)

A1: dep on M2, for use of algebra to show correct conclusion

In the last line of working, \(4(n - 1)\) is used twice (corrected from the printed mark scheme: “\(4(n - 4)\)”).

Higher June 2025 Paper 2R Q13

EdexcelCurrent spec5 marksExpanding Brackets Factorising

(a) Factorise \(4x^2 - 25y^2\) (2)

(b) Show that \(4x(x + 3)(2x - 5)\) can be written in the form \(ax^3 + bx^2 + cx\) where \(a\), \(b\) and \(c\) are integers to be found. (3)

Scheme	Marks
\((2x \pm 5y)(2x \pm 5y)\) or \((2x)^2 - (5y)^2\)	M1
Correct answer only scores full marks (unless from obviously incorrect working) Answer: \((2x + 5y)(2x - 5y)\)	A1
	(2)

Mark scheme (b)

Scheme	Marks
\(4x(x + 3) = 4x^2 + 12x\) or \(4x(2x - 5) = 8x^2 - 20x\) or \((x + 3)(2x - 5) = 2x^2 - 5x + 6x - 15\;(= 2x^2 + x - 15)\)	M1
\((4x^2 + 12x)(2x - 5) = 8x^3 - 20x^2 + 24x^2 - 60x\) or \((8x^2 - 20x)(x + 3) = 8x^3 + 24x^2 - 20x^2 - 60x\) or \(4x(2x^2 - 5x + 6x - 15) = 8x^3 - 20x^2 + 24x^2 - 60x\) or \(4x(2x^2 + x - 15) = 8x^3 + 4x^2 - 60x\)	M1
Working required Answer: \(8x^3 + 4x^2 - 60x\)	A1
	(3)
	(5 marks)

Notes

M1: An expansion with only one error

Do not award this mark for
\(4x^2 + 12x + 8x^2 - 20x\)

M1: ft dep on M1
allow one further error

A1: cao dep on M1
Terms may be in any order but must be simplified

ISW correct factorisation
\(8x^3 + 4x^2 - 60x\) must be seen previously to award 3 marks
eg \(4(2x^3 + x^2 - 15x)\)
\(x(8x^2 + 4x - 60)\)

Do not ISW incorrect simplification or further incorrect work following \(8x^3 + 4x^2 - 60x\)
eg \(8x^3 + 4x^2 - 60x = 2x^3 + x^2 - 15x\) gets M2A0

M2 for 3 terms (out of a maximum of 4 terms) from:
\(8x^3 - 20x^2 + 24x^2 - 60x\)
If not M2, then M1 for 2 correct out of a maximum of 4

Higher June 2025 Paper 2 Q12

EdexcelCurrent spec3 marksExpanding Brackets

12 Expand and simplify \(3x(2x + 5)(7x - 4)\)

(3)

Mark scheme

Scheme	Marks
\(3x(2x + 5) = 6x^2 + 15x\) or \(3x(7x - 4) = 21x^2 - 12x\) or \((2x + 5)(7x - 4) = 14x^2 - 8x + 35x - 20\) \((14x^2 + 27x - 20)\)	M1
\((6x^2 + 15x)(7x - 4) = 42x^3 - 24x^2 + 105x^2 - 60x\) \((21x^2 - 12x)(2x + 5) = 42x^3 + 105x^2 - 24x^2 - 60x\) \(3x(14x^2 - 8x + 35x - 20) = 42x^3 - 24x^2 + 105x^2 - 60x\) \(3x(14x^2 + 27x - 20) = 42x^3 + 81x^2 - 60x\)	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(42x^3 + 81x^2 - 60x\)	A1
	(3)
	(3 marks)

Notes

M1: An expansion with only one error.
Do not award this mark for
\(6x^2 + 15x + 21x^2 - 12x\)
or
\((6x^2 + 15x)(21x^2 - 12x)\)

M1: ft dep on M1

allow one further error

A1: cao (terms may be in any order but must be simplified) dep on M1

ISW correct factorisation eg \(3(14x^3 + 27x^2 - 20x)\)
Do not ISW incorrect simplification eg \(14x^3 + 27x^2 - 20x\)

M2 for 3 (out of a maximum of 4) of \(42x^3 + 105x^2 - 24x^2 - 60x\)
(M1 for 2 correct out of a maximum of 4)

Higher November 2024 Paper 1 Q15

EdexcelCurrent spec3 marksExpanding Brackets

15 Write \(3x(2x - 1)(5x + 4)\) in the form \(ax^3 + bx^2 + cx\) where \(a\), \(b\) and \(c\) are integers.

(3)

Mark scheme

Scheme	Marks
\(3x(2x - 1) = 6x^2 - 3x\) or \(3x(5x + 4) = 15x^2 + 12x\) or \((2x - 1)(5x + 4) = 10x^2 + 8x - 5x - 4\) \((10x^2 + 3x - 4)\)	M1
\((6x^2 - 3x)(5x + 4) = 30x^3 + 24x^2 - 15x^2 - 12x\) \((15x^2 + 12x)(2x - 1) = 30x^3 - 15x^2 + 24x^2 - 12x\) \(3x(10x^2 + 8x - 5x - 4) = 30x^3 + 24x^2 - 15x^2 - 12x\) \(3x(10x^2 + 3x - 4) = 30x^3 + 9x^2 - 12x\)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(30x^3 + 9x^2 - 12x\)	A1
	(3)
	(3 marks)

Notes

M1: An expansion with only one error.

Do not award this mark for
\(6x^2 - 3x + 15x^2 + 12x\)

M1: ft dep on M1

allow one further error

A1: cao (terms may be in any order but must be simplified) dep on M1
accept \(a = 30\), \(b = 9\), \(c = -12\)

ISW correct factorisation
eg
\(3(10x^3 + 3x^2 - 4x)\)
Do not ISW incorrect simplification

M2 for 3 (out of a maximum of 4) of
\(30x^3 + 24x^2 - 15x^2 - 12x\)

(M1 for 2 correct out of a maximum of 4)

Higher November 2024 Paper 1 Q3

EdexcelCurrent spec6 marksExpanding Brackets Solving Simple Equations

(a) Simplify \(\left(p^3\right)^5\) (1)

(b) Expand and simplify \(2n(4n + 3) + n(n - 4)\) (2)

Scheme	Marks
\(p^{15}\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(8n^2 + 6n + n^2 - 4n\)	M1
\(9n^2 + 2n\)	A1
	(2)

Notes

M1: for expanding with at least 3 correct terms (must see for example, \(8n^2\) and not just \(2n \times 4n\))(can assume that no sign in front of a number is a + if terms written in a list or table)

A1: oe \(2n + 9n^2\) or \(n(9n + 2)\) or \(n(2 + 9n)\)

Mark scheme (c)

Scheme	Marks
eg \(2x + 5 = 12 - 3x\) or \(\dfrac{2}{3}x + \dfrac{5}{3} = 4 - x\) oe	M1
eg \(2x + 3x = 12 - 5\) or \(5x = 7\) or \(5 - 12 = -3x - 2x\) or \(-7 = -5x\) or \(\dfrac{2}{3}x + x = 4 - \dfrac{5}{3}\) oe or \(\dfrac{5}{3}x = \dfrac{7}{3}\) oe	M1
Working required Answer: \(\dfrac{7}{5}\)	A1
	(3)
	(6 marks)

Notes

M1: for removal of fraction and multiplying out RHS correctly by 3
or
separating fraction (LHS) in an equation

M1: ft (dep on 4 terms) correctly rearranging their 4 term equation for terms in \(x\) on one side of equation and number terms on the other

A1: oe eg 1.4 or \(1\dfrac{2}{5}\) dep on M2