23 Ben is trying to make \(m\) the subject of \(p = \dfrac{m}{3} + 5\)
Here is his working.
\[\begin{aligned} p - 5 &= \frac{m}{3} \\ 3 \times p - 5 &= m \\ m &= 3p - 5 \end{aligned}\]
Ben’s answer is wrong.
(a) What mistake has Ben made? (1)
(b) Factorise fully \(2x^3y + 4xy^2\) (2)
Mark scheme (a)
Answer
Mark
Mark scheme
Mistake identified
C1
for identifying the mistake
Acceptable examples \(p - 5\) should be multiplied by 3 (\(-\))5 should be multiplied by 3 All of left side / everything should be multiplied by 3 He failed to multiply the 5 as well He should have / didn’t put brackets around the \(p - 5\) (The \(3p - 5\)) should be \(3p - 15\) (The \(-5\)) should be \(-15\) / (the 5) should be 15 (The answer should be) \(m = 3p - 15\) / \(m = 3(p - 5)\) He only times the \(p\) by 3
Not acceptable examples The first line should be \(3p = m + 5\) He should have multiplied everything Ben didn’t divide \(p - 5\) by 3 He failed to multiply the \(p - 5\) He failed to multiply the (\(-\))5 He only times the \(p\) He should have multiplied by 3 first He only multiplied one side by 3 He needs to get rid of the fraction Should have used brackets Just circling the \(3 \times p - 5\) and / or the \(m = 3p - 5\) Needs to multiply the 5 by \(-3\) He should have done \(p - 5 \times 3\)
Mark scheme (b)
Answer
Mark
Mark scheme
\(2xy(x^2 + 2y)\)
B2
for \(2xy(x^2 + 2y)\) oe eg \(2(x^2 + 2y)xy\)
(B1
for \(2x(x^2y + 2y^2)\) or \(2y(x^3 + 2xy)\) or \(xy(2x^2 + 4y)\) or for correctly identifying the HCF in the factorisation of the form \(2xy(ax^2 \pm \ldots)\) or \(2xy(\ldots \pm by)\) where \(a\) and \(b\) are integers or \((x^2 + 2y)\) as a factor eg \(2x(x^2 + 2y)\))
Additional guidance
\(\ldots\) can be numerical or algebraic but not equal to 0 or absent
2 Ben is trying to make \(m\) the subject of \(p = \dfrac{m}{3} + 5\)
Here is his working.
\[\begin{aligned} p - 5 &= \frac{m}{3} \\ 3 \times p - 5 &= m \\ m &= 3p - 5 \end{aligned}\]
Ben’s answer is wrong.
(a) What mistake has Ben made? (1)
(b) Factorise fully \(2x^3y + 4xy^2\) (2)
Mark scheme (a)
Answer
Mark
Mark scheme
Mistake identified
C1
for identifying the mistake
Acceptable examples \(p - 5\) should be multiplied by 3 (\(-\))5 should be multiplied by 3 All of left side / everything should be multiplied by 3 He failed to multiply the 5 as well He should have / didn’t put brackets around the \(p - 5\) (The \(3p - 5\)) should be \(3p - 15\) (The \(-5\)) should be \(-15\) / (the 5) should be 15 (The answer should be) \(m = 3p - 15\) / \(m = 3(p - 5)\) He only times the \(p\) by 3
Not acceptable examples The first line should be \(3p = m + 5\) He should have multiplied everything Ben didn’t divide \(p - 5\) by 3 He failed to multiply the \(p - 5\) He failed to multiply the (\(-\))5 He only times the \(p\) He should have multiplied by 3 first He only multiplied one side by 3 He needs to get rid of the fraction Should have used brackets Just circling the \(3 \times p - 5\) and / or the \(m = 3p - 5\) Needs to multiply the 5 by \(-3\) He should have done \(p - 5 \times 3\)
Mark scheme (b)
Answer
Mark
Mark scheme
\(2xy(x^2 + 2y)\)
B2
for \(2xy(x^2 + 2y)\) oe eg \(2(x^2 + 2y)xy\)
(B1
for \(2x(x^2y + 2y^2)\) or \(2y(x^3 + 2xy)\) or \(xy(2x^2 + 4y)\) or for correctly identifying the HCF in the factorisation of the form \(2xy(ax^2 \pm \ldots)\) or \(2xy(\ldots \pm by)\) where \(a\) and \(b\) are integers or \((x^2 + 2y)\) as a factor eg \(2x(x^2 + 2y)\))
Additional guidance
\(\ldots\) can be numerical or algebraic but not equal to 0 or absent
11 Kate was asked to factorise \(x^2 + 5x + 6\) in the form \((x + a)(x + b)\)
Kate says,
“The sum of \(a\) and \(b\) must be 6 and the product of \(a\) and \(b\) must be 5”
(a) Explain what is wrong with Kate’s statement. (1)
(b) Factorise fully \(2m^2 - 2\) (2)
(c) Factorise fully \(ax + bx - ay - by\) (2)
Mark scheme (a)
Answer
Mark
Mark scheme
Explanation
C1
for explanation Acceptable examples the sum must be 5 and the product must be 6 she had the sum and the product the wrong way round it should be the other way around \(a\) and \(b\) must be multiplied together to make 6
Not acceptable examples the answer should be \((x + 3)(x + 2)\) the product of \(a\) and \(b\) is not 5
Mark scheme (b)
Answer
Mark
Mark scheme
\(2(m - 1)(m + 1)\)
M1
for a correct partial factorisation, eg \(2(m^2 - 1)\) or \((2m - 2)(m + 1)\) or \((m - 1)(2m + 2)\)
A1
cao
Mark scheme (c)
Answer
Mark
Mark scheme
\((a + b)(x - y)\)
M1
for a correct partial factorisation, eg \(x(a + b) - y(a + b)\) or \(x(a + b) + y(-a - b)\) or \(a(x - y) + b(x - y)\)
for factorisation eg \((4 - x^2 =)\ (2 - x)(2 + x)\) and \((x^2 + 3x =)\ x(x + 3)\) or for inversion and multiplication (condone incorrect factorising) eg \(\dfrac{4 - x^2}{x^2 + 3x} \times \dfrac{x + 3}{x + 2}\) oe
M1
for factorisation of both quadratics and inversion and multiplication, eg \(\dfrac{(2 - x)(2 + x)}{x(x + 3)} \times \dfrac{x + 3}{x + 2}\) allow sign errors in one bracket
(c) Make \(g\) the subject of the formula \(\ f = 3g + 11\) (2)
Mark scheme (a)
Answer
Mark
Mark scheme
\(13y - 1\)
M1
for method to expand one bracket or collect like terms eg \(3 \times 2y - 3 \times 5\ (= 6y - 15)\) or \(7 \times y + 7 \times 2\ (= 7y + 14)\) or \(3 \times 2y + 7 \times y\ (= 6y + 7y)\) or \(3 \times -5 + 7 \times 2\ (= -15 + 14)\)
A1
oe
Additional guidance
May be implied by \(13y\) or \(-1\)
Mark scheme (b)
Answer
Mark
Mark scheme
\(3x(2x + 5)\)
B2
oe
(B1
for \(3(2x^2 + 5x)\) or \(x(6x + 15)\) or \(3x(ax + b)\))
Mark scheme (c)
Answer
Mark
Mark scheme
\(g = \dfrac{f - 11}{3}\)
M1
for correct first step to rearrange eg \(f - 11 = 3g + 11 - 11\) or \(f - 11 = 3g\) or eg \(\dfrac{f}{3} = \dfrac{3g}{3} + \dfrac{11}{3}\) or \(-3g = 11 - f\) or answer ambiguously shown, eg \(g = f - 11 \div 3\) or given as \(\dfrac{f - 11}{3}\)
14 Show that \(\dfrac{x^2 - x - 6}{2x^2 - 5x - 3}\) can be written in the form \(\dfrac{ax + b}{cx + d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (3)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{x + 2}{2x + 1}\)
M1
for correctly factorising one expression, eg \((x - 3)(x + 2)\) or \((x - 3)(2x + 1)\)
M1
for factorising both expressions, eg \((x - 3)(x + 2)\) and \((x - 3)(2x + 1)\)
for a correct first step, eg subtracts 3 from both sides or multiplies all terms by 2
M1
(dep M1) for a correct second step, eg multiplies both sides by 2 or divides both sides by 5 or gives the critical value, 6.
A1
for \(x \gt 6\)
Additional guidance
Could be seen as an equation for both method marks, eg \(5x + 6 = 36\) or \(5x = 30\)
First 2 marks may be awarded for critical value of 6, eg \(x = 6\)
Mark scheme (b)
Answer
Mark
Mark scheme
\((x + 9)(x + 1)\)
M1
for \((x \pm 1)(x \pm 9)\) or for \((x + a)(x + b)\) where product of \(a\) and \(b\) = 9, eg \((x + 3)(x + 3)\) or \((x - 3)(x - 3)\) or the sum of \(a\) and \(b\) = 10, eg \((x + 5)(x + 5)\) or \((x + 6)(x + 4)\)
(a) Prove that \[(2m + 1)^2 - (2n - 1)^2 = 4(m + n)(m - n + 1)\] (3)
Sophia says that the result in part (a) shows that the difference of the squares of any two odd numbers must be a multiple of 4
(b) Is Sophia correct? You must give reasons for your answer. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
Proof
M1
for expansion of \((2m + 1)^2\) or \((2n - 1)^2\), all 4 terms correct with or without signs (and no additional terms) or 3 out of 4 terms correct with signs, eg \(4m^2 + 2m + 2m + 1\) or \(4n^2 - 2n - 2n + 1\)
or for correct expansion of \(4(m + n)(m - n + 1)\) or \((m + n)(m - n + 1)\) eg \(4m^2 - 4mn + 4m + 4mn - 4n^2 + 4n\) oe or \(m^2 - mn + m + mn - n^2 + n\) oe
for yes with explanation, eg \(2m + 1\) and \(2n - 1\) are odd numbers (for any positive integer value of \(m\), \(n\)) and the right-hand side is a multiple of 4
for isolating \(x\) terms, eg \(4x = 37 + 7\) or \(4x = 44\) or for \(x - \frac{7}{4} = \frac{37}{4}\) or for \(37 + 7 = 44\) followed by \(\text{``}44\text{''} \div 4\ (= 11)\)
15 Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8 (3)
Mark scheme
Answer
Mark
Mark scheme
proof
C1
for writing an expression for an odd number, eg \(2n + 1\) or \(2n - 1\) (assuming \(n\) is any integer) or states \(n\) is even and eg \((n + 1)\) or \((n + 3)\) as odd numbers
C1
for a correct expression of the form \((2n + 1)^2 - (2n - 1)^2\) expanded eg \(4n^2 + 12n + 9 - (4n^2 + 4n + 1)\) or \(4n^2 + 4n + 1 - (4n^2 - 4n + 1)\) or \((2n + 1 + 2n - 1)(2n + 1 - (2n - 1))\) or when \(n\) is even and eg \((n^2 + 6n + 9) - (n^2 + 2n + 1)\) (\(= 4n + 8\))
C1
for a correct simplified expression as a multiple of 8 eg \(8n + 8\) or \(8n\) or when \(n\) is even and eg \(4n + 8\) and full explanation as to why \(4(n + 2)\) is always a multiple of 8
Additional guidance
Expansion of \((2n - 1)^2 - (2n + 1)^2\) oe is acceptable
(a) Write \(\dfrac{4x^2 - 9}{6x + 9} \times \dfrac{2x}{x^2 - 3x}\) in the form \(\dfrac{ax + b}{cx + d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (3)
(b) Express \(\dfrac{3}{x + 1} + \dfrac{1}{x - 2} - \dfrac{4}{x}\) as a single fraction in its simplest form. (3)
for using ‘\(a\)’ \(= x^2 + 4\) and ‘\(b\)’ \(= x^2 - 2\) OR multiplying out both brackets, at least one fully correct
M1
(dep) for a correct expression for (‘\(a\)’ + ‘\(b\)’)(‘\(a\)’ − ‘\(b\)’) with no additional brackets, simplified or unsimplified eg \((x^2 + 4 + x^2 - 2)(x^2 + 4 - x^2 + 2)\) or \((2x^2 + 2) \times 6\) OR ft for a correct expression without brackets, simplified or unsimplified eg \(x^4 + 8x^2 + 16 - x^4 + 4x^2 - 4\)
A1
for \(12(x^2 + 1)\) or \(12x^2 + 12\) oe
Additional guidance
M1 (first): Correct 4 terms if not simplified or 3 terms if simplified
for \(2a(ab + 3b^2)\) or \(2b(a^2 + 3ab)\) or \(ab(2a + 6b)\) or \(2ab(\)2 term expression with terms in \(a\) or \(b\) or \(ab\), can include constants), eg \(2ab(1a + 3ab)\), \(2ab(1 + 3b)\)
24 Prove that \(\quad \dfrac{60x^4 - 15x^2}{-2x - 1} \times \dfrac{1}{6x - 3} \quad\) can never be positive. [4 marks]
Mark scheme
Answer
Mark
Comments
Partially or fully factorises numerator
M1
eg \(15x(4x^3 - x)\) or \(x^2(60x^2 - 15)\) or \(15x^2(2x - 1)(2x + 1)\)
Factorises at least one denominator or correct multiplication of the denominators
M1
eg \(-(2x + 1)\) and/or \(3(2x - 1)\)
eg \(-12x^2 + (6x - 6x +)\ 3\) may be in a grid
Converts numerators and denominators into terms which can be fully cancelled
M1dep
dep on M1M1 eg \(\dfrac{15x^2(2x - 1)(2x + 1)}{-(2x + 1)3(2x - 1)}\) or \(\dfrac{15x^2(2x - 1)(2x + 1)}{-(2x + 1)} \times \dfrac{1}{3(2x - 1)}\) or \(\dfrac{5x^2(12x^2 - 3)}{-(12x^2 - 3)}\) factorisation and cancelling may be done in stages
\(\dfrac{15x^2}{-3}\) or \(-5x^2\) with M3 awarded and explanation that \(x^2\) cannot be negative
A1
oe with full cancelling of algebraic terms
condone explanation that \(x^2\) must be positive
Additional guidance
\(\dfrac{20x^4 - 5x^2}{(-2x - 1)(2x - 1)}\) or \(\dfrac{20x^4 - 5x^2}{-4x^2 + 1}\) implies M1M1 as 3 has been cancelled from both
M1M1
\(\dfrac{5x^2(2x + 1)}{-(2x + 1)}\) implies complete factorisation of numerator and denominator as 3 and \((2x - 1)\) have been cancelled from both
(b) A sequence has \(n\)th term \(\quad 3n^2 + 5n + 2\)
Are any of the terms in the sequence a prime number?
Tick a box.
Yes
No
Give a reason for your answer. [1 mark]
Mark scheme (a)
Answer
Mark
Comments
\((3n + 2)(n + 1)\)
B2
oe product of brackets any consistent letter condone = 0 ignore any attempt to solve B1 \((3n + 2)\) or \((n + 1)\) seen in a product of 2 linear brackets or \(3n(n + 1) + 2(n + 1)\) or \(n(3n + 2) + (3n + 2)\)
Additional guidance
\((3n + 2)(n + 1) + k\)
B1
Mark scheme (b)
Answer
Mark
Comments
No and valid reason
B1
valid reasons include the sequence is always even and greater than 2 \(n + 1\) and \(3n + 2\) cannot be equal to 1 each term can be made by multiplying (whole) numbers together not equal to 1 \(n + 1\) and \(3n + 2\) are factors not equal to 1
Additional guidance
Yes ticked
B0
No reason given
B0
No ticked, and every term in the sequence is even and the first term is 10
B1
No ticked, and odd + odd + 2 is even, even + even + 2 is even and first term is 10
\(\left(\dfrac{6}{a} =\right) \dfrac{24}{4a}\) or converts both fractions to a common denominator or correct unsimplified fraction eg \(\dfrac{26}{8a}\) or \(\dfrac{13a}{4a^2}\) or \(\dfrac{3.25}{a}\)
M1
oe eg \(\dfrac{48}{8a}\) and \(\dfrac{22}{8a}\) or \(\dfrac{24a}{4a^2}\) and \(\dfrac{11a}{4a^2}\)
\(\dfrac{13}{4a}\)
A1
Additional guidance
Do not ignore further work eg \(\dfrac{13}{4a}\) followed by answer \(\dfrac{3.25}{a}\)
M1A0
Allow a division sign rather than a fraction line for M1 only eg \(26 \div 8a\) eg \(13 \div 4a\)
\(y(y + 7)\) or \(y^2 + 7y\) with no other working
M1M1M1A1
Answer \(\dfrac{y(y + 7)}{1}\) or \(\dfrac{y^2 + 7y}{1}\)
M1M1M1A0
Ignore the consistent use of a different variable within a factorisation
Award SC1 only if there are no correct factorisations eg correct factorisation to \((y + 7)(y + 3)\) and correct expansion to \(y^4 - 3y^3 + 10y^3 - 30y^2 + 21y^2 - 63y\)
\(10cd + 5c\) or \(10dc + 5c\) or \(5c + 10cd\) or \(5c + 10dc\)
B2
B1 fully simplified first term ie \(10cd\) or \(10dc\) or correct expansion not fully simplified eg \(10 \times cd + 5c\) or \(5c \times 2d + 5c\) (\(\times\) 1) or \(5c2d + 5 \times c\)
Additional guidance
Further incorrect work after a B2 response is B1 eg \(10cd + 5c = 15cd\)
B1
Further incorrect work after a B1 response is still B1 eg \(10cd + 1 = 11cd\)
B1
Mark scheme (c)
Answer
Mark
Comments
\(7(3x + 4)\)
B1
Additional guidance
Condone missing final bracket ie \(7(3x + 4\)
B1
Allow multiplying back out to check their answer
Further incorrect work after a correct response is B0 eg \(7(3x + 4) = 7(7x)\)
Condone multiplication signs for B1 but not B2 Condone \(1x\) for \(x\) for B1 but not B2 Condone incorrect algebraic notation for B1 but not B2 eg \(x(x2 + 6)\)
Do not allow further work for B2 but ignore further work for B1 eg \(2x(x + 3) = 2x(3x)\) eg \(x(2x + 6) = x(8x)\)
Do not ignore further incorrect algebraic simplification for B3 \(2a^2 + 15a - 1 = 17a - 1\)
B2
Do not ignore further incorrect algebraic simplification for B2 \(2a + 15a - 1 = 17a - 1 = 16a\) \(2a^2 + 15a - 1 = 17a - 1 = 16a\)
B1
Mark scheme (b)
Answer
Mark
Comments
\(4y(6y - 5)\) or \(-4y(5 - 6y)\)
B2
B1 \(2y(12y - 10)\) or \(-2y(10 - 12y)\) or \(y(24y - 20)\) or \(-y(20 - 24y)\) or \(4(6y^2 - 5y)\) or \(-4(5y - 6y^2)\) or \(2(12y^2 - 10y)\) or \(-2(10y - 12y^2)\)
Additional guidance
Ignore any ‘solutions’ seen eg \(4y(6y - 5)\) in working with 0 and \(\dfrac{5}{6}\) on answer line
B2
Condone \(4y \times (6y - 5)\)
B2
Condone \(y \times (24y - 20)\)
B1
\((4y + 0)(6y - 5)\)
B1
Do not ignore further incorrect algebraic simplification for B2
Correct factorisation of numerator \(2(2x - 4x^2)\) or \(4(x - 2x^2)\) or \(x(4 - 8x)\) or \(2x(2 - 4x)\) or \(4x(1 - 2x)\) or correct factorisation of denominator \(2(6x - 3)\) or \(3(4x - 2)\) or \(6(2x - 1)\) or correct cancelling by 2 throughout \(\dfrac{2x - 4x^2}{6x - 3}\)
M1
oe with negative coefficients
Correct fraction with numerator \(4x(1 - 2x)\) or \(-4x(2x - 1)\) and denominator \(6(2x - 1)\) or \(-6(1 - 2x)\) or \(-\dfrac{4x}{6}\) or \(\dfrac{-4x}{6}\) or \(\dfrac{4x}{-6}\) or \(\dfrac{2x(2 - 4x)}{-3(2 - 4x)}\) or \(\dfrac{2x(2 - 4x)}{3(4x - 2)}\)
M1dep
oe with cancelling of 2 throughout eg \(\dfrac{2x(1 - 2x)}{3(2x - 1)}\) or \(\dfrac{2x(1 - 2x)}{-3(1 - 2x)}\)
\(-\dfrac{2x}{3}\) or \(-\dfrac{2}{3}x\)
A1
allow \(\dfrac{-2x}{3}\) or \(\dfrac{2x}{-3}\)
Additional guidance
Allow multiplication signs up to M1M1
Allow \(-0.\dot{6}\) for \(-\dfrac{2}{3}\)
Do not allow \(-0.66\ldots\) for \(-\dfrac{2}{3}\)
For the first M1 only, allow any correct factorisation seen within multiple attempts
\(\dfrac{8x^2 - 8}{4x + 4} \quad\) simplifies to the form \(\quad ax + b \quad\) where \(a\) and \(b\) are integers. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
Any correct factorisation of the numerator or the denominator
M1
eg \(8(x^2 - 1)\) or \(4(x + 1)\) or \(2(4x^2 - 4)\) or \(2(2x + 2)\) or \(4(2x^2 - 2)\) or \((4x + 4)(2x - 2)\) or \((4x - 4)(2x + 2)\) or \((8x + 8)(x - 1)\) or \((8x - 8)(x + 1)\) or \(-2(-4x^2 + 4)\) does not need to be seen in a fraction may be implied eg \(\dfrac{2x^2 - 2}{x + 1}\) or \(\dfrac{4x^2 - 4}{2x + 2}\)
Correct fraction with a common algebraic factor in the numerator and the denominator
A1
eg \(\dfrac{8(x + 1)(x - 1)}{4(x + 1)}\) or \(\dfrac{2(2x + 2)(2x - 2)}{2(2x + 2)}\) or \(\dfrac{2(x + 1)(x - 1)}{(x + 1)}\) or \(\dfrac{4(x + 1)(2x - 2)}{4(x + 1)}\) or \(\dfrac{(4x + 4)(2x - 2)}{4x + 4}\)
\(2x - 2\) or \(a = 2\) and \(b = -2\) with M1A1 scored
A1
Alternative method 2
\(4ax^2 + 4ax + 4bx + 4b\)
M1
oe expands \((ax + b)(4x + 4)\) to 4 terms with at least 3 terms correct
Any 2 of \(4a = 8 \qquad 4b = -8 \qquad 4a + 4b = 0\)
A1
\(a = 2\) and \(b = -2\) and shows that third equation is satisfied with M1A1 scored
A1
Additional guidance
M1 is implied by the first A1 eg \(\dfrac{8(x + 1)(x - 1)}{4(x + 1)}\)
M1A1
\(1(8x^2 - 8)\) or \(-1(8 - 8x^2)\) etc
M0
\(2x - 2\) without M1A1 scored
M0A0A0
M1A1 scored and \(2x - 2\) followed by attempt to solve \(2x - 2 = 0\)
M1A1A1
M1A1 scored and \(2x - 2\) followed by \(2(x - 1)\)
M1A1A1
M1A1 scored followed by \(2(x - 1)\) but \(2x - 2\) not seen
B1 \((3x + a)(x + b)\) where \(ab = 7\) or \(a + 3b = -22\) or \((a - 3x)(b - x)\) where \(ab = 7\) or \(a + 3b = 22\)
Additional guidance
\((3x + 1)(x + 7)\)
B1
\((3x - 1)(x - 7)\)
B1
\((3x - 4)(x - 6)\)
B1
\((7 - 3x)(1 - x)\)
B1
\((10 - 3x)(4 - x)\)
B1
\((3x - 1) \times (x - 7)\)
B2
Ignore any ‘solutions’ seen eg \((3x - 1)(x - 7)\) in working with \(\dfrac{1}{3}\) and 7 on answer line
B2
Notes
The second guidance row is printed this way in the published mark scheme. The correct answer \((3x - 1)(x - 7)\) scores B2, as shown in the main table.
