Algebraic Fractions

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Higher June 2025 Paper 2 Q23

EdexcelCurrent spec4 marksAlgebraic Fractions Factorising

23 \(\dfrac{4x^2 - 4x - 120}{5x^2 - 180} \div \dfrac{x^2 + 5x}{10x^2 + 60x} = p\) where \(p\) is an integer.

Find the value of \(p\)

Show clear algebraic working.

(4)

Mark scheme

Scheme	Marks
\(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \div \dfrac{x(x + 5)}{10x(x + 6)}\;(= p)\) or \(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \times \dfrac{10x(x + 6)}{x(x + 5)}\;(= p)\) or 2 from \(4(x - 6)(x + 5)\) or \(5(x - 6)(x + 6)\) or \(x(x + 5)\) or \(10x(x + 6)\)	M1
\(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \div \dfrac{x(x + 5)}{10x(x + 6)}\;(= p)\) or \(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \times \dfrac{10x(x + 6)}{x(x + 5)}\;(= p)\) or \(4(x - 6)(x + 5)\) and \(5(x - 6)(x + 6)\) and \(x(x + 5)\) and \(10x(x + 6)\)	M1
\(\dfrac{4x^2 - 4x - 120}{5x^2 - 180} \times \dfrac{10x^2 + 60x}{x^2 + 5x}\;(= p)\)	M1
Working required Answer: 8	A1
	(4)
	(4 marks)

Notes

M1: for factorising 2 or 3 of the quadratics fully – could be implied by 2 factors cancelled correctly

NB factors must be in the form \((ax + b)\)

NB Substitution of values of \(x\) into the given equation is not an acceptable algebraic method

M1: for factorising all of the quadratics fully – could be implied by 2 factors cancelled correctly

NB factors must be in the form \((ax + b)\)

M1: for inverting the 2^nd fraction (this mark can be awarded at any time and may be awarded with incorrect factorisation if meaning is clear)

A1: oe dep on M3

ALT

Scheme	Marks
\(\dfrac{40x^4 + 200x^3 - 1440x^2 - 7200x}{5x^4 + 25x^3 - 180x^2 - 900x}\)	M3
8	A1

Notes

M3: for a correct expression (no errors)

A1: oe dep on M3

Higher June 2025 Paper 1 Q12

EdexcelCurrent spec3 marksAlgebraic Fractions

12 Express \(\dfrac{5}{4} + \dfrac{x - 3}{6x}\) as a single fraction in its simplest form.

(3)

Mark scheme

Scheme	Marks
eg \(\dfrac{5(6x)}{24x} + \dfrac{4(x - 3)}{24x}\) oe or \(\dfrac{5(6x)}{4(6x)} + \dfrac{4(x - 3)}{4(6x)}\) oe or \(\dfrac{30x}{24x} + \dfrac{4(x - 3)}{24x}\) oe or \(\dfrac{30x + 4(x - 3)}{24x}\) oe or \(\dfrac{15x}{12x} + \dfrac{2(x - 3)}{12x}\) oe or \(\dfrac{15x + 2(x - 3)}{12x}\) oe	M1
eg \(\dfrac{30x + 4x - 12}{24x}\) oe or \(\dfrac{30x}{24x} + \dfrac{4x - 12}{24x}\) oe or \(\dfrac{30x}{24x} + \dfrac{4x}{24x} - \dfrac{12}{24x}\) oe or \(\dfrac{34x}{24x} - \dfrac{12}{24x}\) oe or \(\dfrac{34x - 12}{24x}\) oe or \(\dfrac{15x + 2x - 6}{12x}\) oe	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{17x - 6}{12x}\)	A1
	(3)
	(3 marks)

Notes

M1: for two correct fractions with common denominator
or
a single correct fraction

M1: for correct fraction(s) with bracket(s) expanded correctly

A1: oe but must be simplified eg \(\dfrac{-6 + 17x}{12x}\)

do not ISW incorrect simplification eg \(\dfrac{17x - 6}{12x} = \dfrac{11}{12}\) is M2A0

Higher November 2024 Paper 2 Q23

EdexcelCurrent spec4 marksAlgebraic Fractions Factorising

23 Show that \(\dfrac{16x^2 - 36}{x - 7} \div \dfrac{2x^2 + 7x + 6}{x^2 - 5x - 14} - (7 + 8x) = n\)

where \(n\) is an integer to be found.
Show clear algebraic working.

(4)

Mark scheme

Scheme	Marks
\(16x^2 - 36 = (4x - 6)(4x + 6)\;[= 4(2x + 3)(2x - 3)]\) oe or \(16x^2 - 36 = (8x - 12)(2x + 3)\;[= 4(2x + 3)(2x - 3)]\) oe	M1indep
\(2x^2 + 7x + 6 = (2x + 3)(x + 2)\) and \(x^2 - 5x - 14 = (x - 7)(x + 2)\) [We will make an exception of \(2x^2 + 7x + 6 = (4x + 6)(0.5x + 1)\) as this then cancels with \((4x + 6)\)]	M1indep
\(4(2x - 3) - (7 + 8x)\;(= n)\) [allow invisible brackets ie \(8x - 12 - 7 + 8x\)] or an expression that clearly shows the numerator is – 19 times the denominator eg \(\dfrac{-38x - 57}{2x + 3}\;(= n)\)	M1
Working required Answer: –19	A1
	(4)
	(4 marks)

Notes

M1indep: [NB: the two fractions when divided and cancelled give an answer of \(4(2x - 3)\) or \(2(4x - 6)\) or \(8x - 12\) (any one of these gain 2 marks) ]

M2 for any fraction with completely simplified non-linear numerator and non-linear denominator that will cancel to –19

M1: a linear expression that should give the correct value for \(n\)
(this mark implies previous M marks as not all factorising is necessary)

A1: dep on M2