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 \((4y + 10)(2y - 5) = 14y(2y - 5)\)
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
Numerator of \(6x - 9 - 1 + 3x\) for denominator of \(3x\) or numerator of \(6x^2 - 9x - x + 3x^2\) for denominator of \(3x^2\)
M1dep
oe – will depend on the denominator but all numerator terms must be correct for the denominator used. could be a single fraction or separate fractions with the same denominator
\(\dfrac{9x - 10}{3x}\)
A1
\(3 - \dfrac{10}{3x}\) is not a single fraction so is A0
Alternative method 1 Processes the brackets then divides
\(\dfrac{5x}{10} + \dfrac{6x}{10}\)
M1
oe valid common denominator with both numerators correct eg \(\dfrac{10x}{20} + \dfrac{12x}{20}\)
\(\dfrac{11x}{10}\)
A1
oe single term eg \(\dfrac{22x}{20}\) or \(1.1x\) may be implied eg by single term with roots evaluated that is equivalent to \(\dfrac{11}{5x^2}\)
\(\dfrac{x^{6 \div 2}}{2}\) or \(\dfrac{x^3}{2}\)
M1
may be implied eg by multiplication by \(\dfrac{2}{x^3}\)
their \(\dfrac{11x}{10} \times \dfrac{2}{x^3}\) or \(\dfrac{22x}{10x^3}\) or \(\dfrac{22}{10x^2}\) or \(\dfrac{11x}{5x^3}\) or \(\dfrac{22}{10}x^{-2}\)
M1dep
oe multiplication eg \(\dfrac{11x}{10} \times 2x^{-3}\) their \(\dfrac{11x}{10}\) can be unprocessed dep on 2nd M1
\(\dfrac{11}{5x^2}\) or \(\dfrac{11}{5}x^{-2}\) or \(2.2x^{-2}\)
A1
allow \(2\dfrac{1}{5}x^{-2}\) or \(\dfrac{2.2}{x^2}\)
Alternative method 2 Divides then expands the brackets
\(\dfrac{x^{6 \div 2}}{2}\) or \(\dfrac{x^3}{2}\)
M1
may be implied eg by multiplication by \(\dfrac{2}{x^3}\)
\(\dfrac{2x}{2x^3} + \dfrac{6x}{5x^3}\) or \(\dfrac{1}{x^2} + \dfrac{6}{5x^2}\)
M1dep
oe expansion of brackets
\(\dfrac{10x}{10x^3} + \dfrac{12x}{10x^3}\) or \(\dfrac{5}{5x^2} + \dfrac{6}{5x^2}\) or \(\dfrac{22x}{10x^3}\) or \(\dfrac{22}{10x^2}\) or \(\dfrac{11x}{5x^3}\) or \(\dfrac{22}{10}x^{-2}\)
M1dep
oe valid common denominator with both numerators correct eg \(\dfrac{10x^4}{10x^6} + \dfrac{12x^4}{10x^6}\) or \(\dfrac{22x^4}{10x^6}\) roots must be processed
\(\dfrac{11}{5x^2}\) or \(\dfrac{11}{5}x^{-2}\) or \(2.2x^{-2}\)
A1
allow \(2\dfrac{1}{5}x^{-2}\) or \(\dfrac{2.2}{x^2}\)
Additional guidance
Any single fraction with roots evaluated that is equivalent to \(\dfrac{11}{5x^2}\)
4 marks
Allow inclusion of \(\pm\) from the square root for up to 4 marks
\(\dfrac{11}{5x^2}\) in working with answer \(\dfrac{11}{5}x^2\)
4 marks
Alt 1 \(\dfrac{11x}{10}\) subsequently squared and not recovered
2 Write \(\quad \dfrac{5}{6a} + \dfrac{a}{4} \quad\) as a single fraction.
Give your answer in its simplest form. [2 marks]
Mark scheme
Answer
Mark
Comments
Both fractions written with a common denominator (could be written as a single fraction) which is a multiple of \(6a\) and 4 with at least one correct (term of the) numerator
M1
oe eg \(\ \dfrac{20}{24a}\ \) or \(\ \dfrac{6a^2}{24a}\ \) or \(\ \dfrac{4(5)}{4(6a)}\) or \(\ \dfrac{20+6a^2}{24a}\) allow decimals in fraction eg \(\ \dfrac{5+1.5a^2}{6a}\)
\(\dfrac{10+3a^2}{12a}\)
A1
Additional guidance
Penalise further working
\(\dfrac{10+3a^2}{12}\ \) is likely to come from correct working
(a) Write \(\quad \dfrac{7}{9x} + \dfrac{2}{3x^2} \quad\) as a single fraction in its simplest form. [3 marks]
(b) Show that \(\qquad \dfrac{x^4}{x + 4} \times \dfrac{x + 2}{x} \div \dfrac{x^2}{3x + 12}\)
simplifies to the form \(\ ax^2 + bx \ \) where \(a\) and \(b\) are integers. [4 marks]
Mark scheme (a)
Answer
Mark
Comments
Valid common denominator with at least one numerator correct
M1
eg \(\dfrac{7x}{9x^2}\) and \(\dfrac{\mathrm{a}}{9x^2}\) or \(\dfrac{7x + \mathrm{a}}{9x^2}\) or \(\dfrac{\mathrm{b}}{9x \times 3x^2}\) and \(\dfrac{2 \times 9x}{9x \times 3x^2}\) numerators and denominators may be seen as products a can be numerical or algebraic b can be numerical or algebraic
Valid common denominator with both numerators correct
M1dep
\(\dfrac{7x}{9x^2}\) and \(\dfrac{6}{9x^2}\) or \(\dfrac{7 \times 3x^2}{9x \times 3x^2}\) and \(\dfrac{2 \times 9x}{9x \times 3x^2}\) numerators and denominators may be seen as products
\(\dfrac{7x + 6}{9x^2}\) or \(\dfrac{7x + 6}{(3x)^2}\) with no further work
A1
Additional guidance
\(\dfrac{21x^2 + 18x}{27x^3}\) or \(\dfrac{21x + 18}{27x^2}\) or \(\dfrac{7x^2 + 6x}{9x^3}\)
M2A0
\(\dfrac{7x^{-1} + 6x^{-2}}{9}\)
M2A0
\(7x + 6 / 9x^2\)
M2A0
Mark scheme (b)
Answer
Mark
Comments
Changes division to multiplication and inverts to \(\dfrac{3x + 12}{x^2}\)
M1
may be implied
\((3x + 12 =)\ 3(x + 4)\)
M1
may be implied
Correct expression written as a single fraction or a product must have factor \((x + 4)\) in a numerator and denominator \(x + 4\) or correct expression written as a single fraction or a product must have denominator \(x^3\) or \(x^2\) or \(x\) or 1
A1
may be implied by final A1 eg \(\ \dfrac{3x(x + 2)(x + 4)}{x + 4}\) or \(\dfrac{(3x^2 + 6x)(x + 4)}{x + 4}\) or \(\dfrac{x}{x + 4} \times \dfrac{x + 2}{1} \times 3(x + 4)\) or \(\dfrac{x}{x + 4} \times 3(x + 2)(x + 4)\) or \(\dfrac{3x^4(x + 2)}{x^3}\) or \(x^4 \times \dfrac{x + 2}{x} \times \dfrac{3}{x^2}\) or \(\dfrac{(x + 2)}{x^3} \times 3x^4\) or \(\dfrac{3x^3(x + 2)}{x^2}\) or \(\dfrac{3x^2(x + 2)}{x}\) or \(\dfrac{3x(x + 2)}{1}\) or \(x \times (x + 2) \times 3\) or \(3x(x + 2)\)
\(3x^2 + 6x\)
A1
SC2 \(\ \dfrac{x(x + 2)(3x + 12)}{x + 4}\)
Additional guidance
The list of examples in the first A1 is not exhaustive
10 Rearrange \(\qquad \dfrac{1}{xy} = 4 - \dfrac{3}{y} \qquad\) to make \(x\) the subject. [3 marks]
Mark scheme
Answer
Mark
Comments
A correct first step using algebra
M1
Here are some of the possible alternatives \(\dfrac{1}{x} = y\left(4 - \dfrac{3}{y}\right)\) multiplying through by \(y\) \(1 = xy\left(4 - \dfrac{3}{y}\right)\) multiplying through by \(xy\) \(1 = 4xy - \dfrac{3xy}{y}\) multiplying through by \(xy\) \(y = 4xy^2 - 3xy\) multiplying through by \(xy^2\) \(\dfrac{1}{xy} = \dfrac{4y - 3}{y}\) making the RHS an algebraic fraction \(\dfrac{1 + 3x}{xy} = 4\) rearranging and making the LHS an algebraic fraction
Further correct algebra which leads to an equation that is one step from the final answer.
M1dep
Following two of the above alternatives … \(y = 4xy^2 - 3xy\) \(y = x(4y^2 - 3y)\) M1dep gained
There are many ways of scoring the first M mark. They do not need to give any reasons but you need to check that what they do is valid.
For the M1dep mark you must check that their algebra is correct and will lead to a result that is one step from the final answer. ‘One step from …’ means that when they divide through, they have a correct version where \(x\) is the subject.
Some of the final answers are more compact than others, but we didn’t ask for any simplification so we have to accept a correct answer in any form.
… and, finally, one to look out for … correct answer from wrong working … 0 marks \(\dfrac{1}{xy} = 4 - \dfrac{3}{y} \ \rightarrow\ xy = \dfrac{1}{4} - \dfrac{y}{3} \ \rightarrow\ x = \dfrac{1}{4y} - \dfrac{1}{3} \ \rightarrow\ x = \dfrac{1}{4y - 3}\) (creative thinking !)
oe eg \(\dfrac{1}{4} \pm \sqrt{\dfrac{1}{16}}\) may be implied by both correct solutions
0.5 \(\quad\) 0 or \(\dfrac{1}{2} \quad 0\)
A1
oe
Additional guidance
0.5 \(\quad\) 0 in working but only one of these on answer line
M2A0
Equation left in terms of \(p\) and \(m\)
Zero
Answers only of 0.5 \(\quad\) 0
M2A1
Answer only of 0.5
Zero
Answer only of 0
Zero
If using formula with an error seen the maximum mark is M1M0A0 eg \(\ 2m^2 - m = 0 \quad \dfrac{--1 \pm \sqrt{(-1)^2 - 4 \times 1 \times 0}}{2 \times 2}\) Answers 0.5 \(\quad\) 0
Only needs to be seen once eg1 \(\dfrac{8a}{3a + 6} \times \dfrac{5(a + 2)}{3a^2} \div \dfrac{4}{15a^3}\) eg2 \(\dfrac{8a}{3(a + 2)} \times \dfrac{5a + 10}{3a^2} \times \dfrac{15a^3}{4}\)
Award M2 for fully correct unprocessed expression with full cancelling seen eg \(\dfrac{{}^{2}\cancel{8}a}{3\cancel{(a + 2)}} \times \dfrac{5\cancel{(a + 2)}}{\cancel{3}\,\cancel{a^2}} \times \dfrac{{}^{5}\cancel{15}a^{\cancel{3}\,1}}{\cancel{4}}\) or \(\dfrac{2a}{3} \times 5 \times 5a\) oe
\(\dfrac{50a^2}{3}\) or \(16\dfrac{2}{3}a^2\) or \(16.\dot{6}a^2\)
A1
Additional guidance
\(\dfrac{50 \times a \times a}{3}\)
M2A0
A correct single fraction with \((a + 2)\) cancelled will be M2 eg1 \(\dfrac{250a^2}{15} \qquad\) eg2 \(\dfrac{50a^4}{3a^2}\)
Either ... follow the LHS of the mark scheme for the first three steps Or ... follow the RHS ... do not mix expressions ... the numerators and denominators must match
Correct answer followed by incorrect further work However, \(\dfrac{1}{(2x + 3)(2x + 1)}\) with an attempt to expand the denominator should not be penalised even if their expansion is not correct
oe Correct use of quadratic formula for their 3-term quadratic eg \((-20)^2\) can be \(20^2\) or correct factorisation of their 3-term quadratic or attempt to complete the square for their 3-term quadratic Must be correct up to form \((x - a)^2 + b = c\) or \(k[(x - d)^2 + e] = f\)
1.23 and 2.77
A1
Must be 3 significant figures
Additional guidance
For A1, the word ‘and’ is not needed eg 1.23, 2.77 (with method seen)
M5 A1
Brackets may be recovered throughout
5th M1 may be implied by solutions of their quadratic equation seen
M0 M0 M0 M0 M1 A0 is possible if they have a 3-term quadratic equation
\(\dfrac{x - 2}{x + 3}\) 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
