oe eg \(\dfrac{14x^4 - 28x^3 - 42x^2}{4x^3 - 8x^2 - 12x}\) where fractions have been multiplied and combined into a single fraction
Numerator or Denominator further factorised correctly as a step towards the final answer
M1
eg \(14x^2\left(x^2 - 2x - 3\right)\) or \(14x^2(x - 3)(x + 1)\) or \(2x\left(2x^2 - 4x - 6\right)\) or \(4x\left(x^2 - 2x - 3\right)\) or \(4x(x - 3)(x + 1)\) etc could be implied by cancellation of a common factor neither \(7x^2\left(2x^2 - 4x - 6\right)\) nor \(\left(x^2 - 3x\right)\left(4x + 4\right)\) would score this M mark
Valid common denominator with at least one numerator correct
M1
eg \(\dfrac{9}{12w}\) and \(\dfrac{2}{12w}\) or \(\dfrac{18w}{24w^2}\) and \(\dfrac{4w}{24w^2}\) implied by a correct single fraction not in simplest form eg \(\dfrac{9 + 2}{12w}\) or \(\dfrac{22}{24w}\) or \(\dfrac{22w}{24w^2}\) for common denominator \(4w\) or \(6w\) both numerators must be correct
\(\dfrac{11}{12w}\)
A1
accept \(\dfrac{11}{12}w^{-1}\)
Additional guidance
\(\dfrac{11}{12w}\) followed by incorrect further work
For M1 accept numerator(s) and/or denominator(s) seen as products eg \(\dfrac{6w \times 3}{4w \times 6w}\) and \(\dfrac{4w}{4w \times 6w}\)
M1
\(\dfrac{1}{w}\left(\dfrac{3}{4} + \dfrac{1}{6}\right)\) (does not score M1 at this stage) \(\dfrac{1}{w}\left(\dfrac{9}{12} + \dfrac{2}{12}\right)\)
M1
Omitting \(w\) in working with recovery to answer \(\dfrac{11}{12w}\)
M1A1
Omitting \(w\) in working with no or incorrect answer
M0A0
Mark scheme (b)
Answer
Mark
Comments
\((x - 1)^8(2x + 3)\)
B2
B1 factorised expression with \((x - 1)^8\) as one of the factors allow multiplication signs for B2 or B1
Additional guidance
Condone missing final bracket for B2 or B1 unless final factor is \((x - 1)^8\) eg1 \((x - 1)^8(2x + 3\) eg2 \((2x + 3)(x - 1^8\)
B2 B0
Ignore expansion of a B2 or B1 response
\((x - 1)^8(3 + 2x)\) etc
B2
B2 response followed by further, incorrect work is B1 eg \((x - 1)^8(2x + 3) = 5x(x - 1)^8\)
B1
B1 response followed by further, incorrect work is B1 eg \((x - 1)^8(3x + 4) = 7x(x - 1)^8\)
B1
Ignore any attempt to ‘solve’
\((x - 1)^8(x - 1) + (x - 1)^8(x + 4)\)
B0
\((x - 1)^8(x - 1) + (x + 4)\)
B0
Mark scheme (c)
Answer
Mark
Comments
\(2(2y + 5)(2y - 5)\)
B2
B1 \(2(4y^2 - 25)\) or \((4y + 10)(2y - 5)\) or \((4y - 10)(2y + 5)\) or correct use of difference of two squares eg \(8(y + 2.5)(y - 2.5)\) or \(\dfrac{1}{2}(4y + 10)(4y - 10)\) or \(\left(\sqrt{8}\,y + \sqrt{50}\right)\left(\sqrt{8}\,y - \sqrt{50}\right)\) or \(\left(2\sqrt{2}\,y + 5\sqrt{2}\right)\left(2\sqrt{2}\,y - 5\sqrt{2}\right)\) allow multiplication signs for B2 or B1
Additional guidance
All B1 responses must be equivalent to \(8y^2 - 50\)
Condone missing final bracket for B2 or B1
Ignore expansion of a B2 or B1 response
\(2(5 + 2y)(2y - 5)\) etc
B2
\((10 + 4y)(2y - 5)\) etc
B1
B2 response followed by further, incorrect work is B1 eg \(2(2y + 5)(2y - 5) = (4y + 10)(4y - 10)\)
B1
B1 response followed by further, incorrect work is B1 eg \((4y + 10)(2y - 5) = 14y(2y - 5)\)
oe simplification eg \(-5 \pm \dfrac{\sqrt{68}}{2}\) dep on second M mark only implied by final A1
\((x =)\ 0,\ -5 \pm \sqrt{17}\)
A1
could be written as two separate surds.
Alternative method 2: completes the square
\(x(x^2 + 10x + 8)\ \ (= 0)\) or \(x = 0\)
M1
\(x\left[(x + 5)^2 \ldots\ldots\right]\)
M1
oe eg may not include the common factor of \(x\) at this stage must be from correct quadratic
\(x\left[(x + 5)^2 - 17\right]\)
A1
oe eg \((x + 5)^2 = 17\) dep on second M mark only condone common factor of \(x\) not being there at this stage
\((x =)\ 0,\ -5 \pm \sqrt{17}\)
A1
could be written as two separate surds.
Additional guidance
If \(x = 0\) stated as final answer then it doesn’t need to appear in the working
If \(x = 0\) not stated on answer line then second A mark cannot be awarded: eg \(x = 0\) is seen early in the working with everything else correct \(x = 0\) or \(x(x^2 + 10x + 8)\) not seen in the working but with everything else correct
M2A1A0 M0M1A1A0
Factor theorem used correctly to get complete answer
17 Show that \(\quad \dfrac{21x}{3x^2 - 2x - 8} - \dfrac{7}{x - 2} \quad\) simplifies to \(\quad \dfrac{k}{3x^2 - 2x - 8} \quad\) where \(k\) is an integer. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
\((3x + 4)(x - 2)\)
M1
oe for correct factorisation
\(\dfrac{21x - 7(3x + 4)}{(3x + 4)(x - 2)}\)
M1dep
oe fractions can be written separately denominator must be \((3x + 4)(x - 2)\) or \(3x^2 - 2x - 8\)
oe must be one fraction and brackets expanded in the numerator (could have been simplified and factorised) eg \(\dfrac{-28(x - 2)}{(3x^2 - 2x - 8)(x - 2)}\) expanded denominator is \(3x^3 - 8x^2 - 4x + 16\)
\(\dfrac{-28}{3x^2 - 2x - 8}\)
A1
Additional guidance
Students may use a combination of methods which are covered by oe
\(k = -28\) Could come from using identities or substitutions
SC1
The minus can be written before the fraction rather than with the 28
\(x(2x^2 - 9x - 5)\) or \((2x^2 + x)(x - 5)\) or \((2x + 1)(x^2 - 5x)\) or states \(x = 0\) and \((2x^2 - 9x - 5)\)
M1
\(x(2x + 1)(x - 5)\) or states \(x = 0\) and \((2x + 1)(x - 5)\)
M1dep
do not accept \(x(2x + 1)(2x - 10)\) unless recovered by correct solutions
\((x =)\ 0,\ 5,\ -0.5\)
A1
oe but must come from factorising in M1dep
Additional guidance
\((2x + 1)(x - 5)\) if no other marks scored
SC1
\((x =)\ 0,\ 5,\ -0.5\) without working or by using factor theorem to find solutions without writing it as a product of factors or by using quadratic formula (must be all 3 correct answers)
B1 for any factor taken out correctly. \(3(2x^2y + 7xy)\) or \(x(6xy + 21y)\) or \(y(6x^2 + 21x)\) or \(3x(2xy + 7y)\) or \(3y(2x^2 + 7x)\) or \(xy(6x + 21)\) B1 for correct highest common factor with one term incorrect in the bracket \(3xy(a + 7)\) or \(3xy(2x + b)\)
Additional guidance
Condone final bracket missing
Condone multiplication signs
Condone changing order of terms eg \(3(2x + 7)xy\)
\((3xy + 0)(2x + 7)\) or \((3xy + \quad)(2x + 7)\) or \(3xy\) and \((2x + 7)\) seen but not together
17 Show that \(\quad (x + 1)(x + 3)(x + 4) - x(x^2 + 7x + 11)\)
can be written in the form \(\quad (x + a)(x + b) \quad\) where \(a\) and \(b\) are positive integers. [5 marks]
Mark scheme
Answer
Mark
Comments
\(x^2 + 3x + x + 3\) with three terms correct or \(x^2 + 4x + k\) where \(k\) is a non-zero constant
M1
oe expansion attempt of one pair of brackets eg1 \(\ x^2 + 4x + 3x + 12\) with three terms correct or \(x^2 + 7x + k\) where \(k\) is a non-zero constant
eg2 \(\ x^2 + 4x + x + 4\) with three terms correct or \(x^2 + 5x + k\) where \(k\) is a non-zero constant
\(x^3 + 3x^2 + x^2 + 3x\) or \(x^3 + 4x^2 + 3x\) or \(4x^2 + 12x + 4x + 12\) or \(4x^2 + 16x + 12\)
M1dep
attempt at a full expansion with correct multiplication of their 3 or 4 terms by one of the terms in the remaining bracket oe eg \(x^3 + 4x^2 + 3x^2 + 12x\) or \(x^3 + 7x^2 + 12x\) or \(x^2 + 4x + 3x + 12\) or \(x^2 + 7x + 12\) (\(x^2 + 7x + 12\) must be from an attempt at a full expansion) or \(x^3 + 4x^2 + x^2 + 4x\) or \(x^3 + 5x^2 + 4x\) or \(3x^2 + 12x + 3x + 12\) or \(3x^2 + 15x + 12\)
ft M2A0 full simplification of their \((x^3 + 8x^2 + 19x + 12) - x^3 - 7x^2 - 11x\) their \((x^3 + 8x^2 + 19x + 12)\) must be a cubic
\(x^2 + 8x + 12\) and \((x + 6)(x + 2)\) or \((x + 2)(x + 6)\)
A1
oe product of brackets
Additional guidance
1st M1 Do not allow omissions or extras eg1 \(\ x^2 + 3x + 3\) eg2 \(\ x^2 + 3x + x + 3 + x^2\)
M0 M0
For the first 2 marks terms may be seen in a grid
If 1st A1 has been awarded with terms not collected, A1ft can still be awarded using their simplified cubic eg \(\ x^3 + 4x^2 + 3x + 4x^2 + 16x + 12\) \(= x^3 + 8x^2 + 18x + 12\) \(x^3 + 8x^2 + 18x + 12 - x^3 - 7x^2 - 11x\) \(= x^2 + 7x + 12\)
M1M1A1
A1ftA0
First A1 may be seen embedded eg \(\ x^3 + 8x^2 + 19x + 12 - x^3 - 7x^2 - 11x\)
M1M1A1
If an attempt at the expansion of all three brackets in one go is made it must be fully correct to gain M2A1, otherwise M0M0A0 eg \(\ x^2 + 3x + x + 3 + x^2 + 4x\)
M0M0A0
Allow recovery of missing brackets when subtracting \(x^3 + 7x^2 + 11x\) from their cubic
For final A1 allow \(x^2 + 8x + 12\) and \(a = 6 \ \ b = 2\) or \(x^2 + 8x + 12\) and \(a = 2 \ \ b = 6\)
Ignore equating to zero and/or any ‘solving’ of an equation
B1 \((p + 6)^{10}(p + 6 - 1)\) or states \(x = p + 6\) and \(x^{10}(x - 1)\) (any letter for \(x\) other than \(p\)) or correct partial factorisation eg \((p + 6)[(p + 6)^{10} - (p + 6)^9]\) or \((p + 6)^2[(p + 6)^9 - (p + 6)^8]\)
Additional guidance
Any shape of bracket may be used
\((p + 6)^{10}((p + 6) - 1)\)
B1
Missing brackets must be recovered eg \(\ p + 5\,(p + 6)^{10}\) not recovered and B1 response not seen
B0
Condone \(\ (p + 6)^{10}(p + 5\)
B2
Condone \(\ (p + 6)^{10}(p + 6 - 1\)
B1
\((p + 6)^{10}(p + 5)\) followed by expansion attempt
B1 correct factorised expression with a common factor involving at least two variables eg \(pq(12q^2r - 18qr^2 + 24qr)\) or \(2q^2r\,(6pq - 9pr + 12p)\) or common factor \(6pq^2r\) with two out of the three terms in the bracket correct eg \(6pq^2r\,(2q - 3r + 4p)\)
Additional guidance
B2 answer followed by further work
B1
\(6pq^2r\,(2q - 3r + 4)\) in working with \(6qp^2r\,(2q - 3r + 4)\) on answer line
B1
B1 answer followed by further work
B1
\(2q^2r\,(6pq - 9pr + 12p)\) in working with \(2p^2r\,(6pq - 9pr + 12p)\) on answer line
B1
Use of multiplication signs scores a maximum of B1
\(qpq(12qr - 18r^2 + 24r)\)
B1
\(6pqrq\,(2q - 3r + 4)\)
B1
Mark scheme (b)
Answer
Mark
Comments
Correct factorised expression with a common factor
\(\dfrac{x - 2}{x + 3}\) or \(\dfrac{2 - x}{-x - 3}\) from incorrect method
M0 A0
Allow fraction with correct factorisations and common factors crossed out
M1 A1
Allow \(\ x - 2 \,/\, x + 3\)
M1 A1
Mark scheme (b)
Answer
Mark
Comments
\(w^2x^3y^2(w^3 + x^3y)\)
B2
B1 A correct partial factorisation with at least one variable fully factorised eg1 \(y^2(w^5x^3 + w^2x^6y)\) eg2 \(w^2x(w^3x^2y^2 + x^5y^3)\) or a correct partial factorisation with all three variables as factors eg \(wxy(w^4x^2y + wx^5y^2)\) or full common factor with one term in brackets correct \(w^2x^3y^2(w^3 + \ldots)\) or \(w^2x^3y^2(\ldots + x^3y)\) Must be two terms in each bracket
Additional guidance
Allow multiplication signs and 1s and unnecessary brackets for B2 and B1
Use of negative or fractional or zero powers is a maximum of B1
\(w^2x^3y^2(w^3 + x^3y)\) followed by incorrect further work
B1
Answer line incorrect, check working lines for possible B1
