Download Simplifying - Expanding brackets

Simplifying - Expanding brackets
113 minutes
106 marks
Page 1 of 28
Q1.
Expand
w(w + 6)
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Answer ......................................................................
(Total 2 marks)
Q2.
(a)
Expand
3(x − 6)
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Answer ......................................................................
(1)
(b)
Factorise
5y − 10
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Answer ......................................................................
(1)
(c)
3(4w + 1) − 5(3w − 2)
Expand and simplify
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Answer ......................................................................
(3)
(Total 5 marks)
Q3.
Expand and simplify
5(x − 3) − 2(x − 1)
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Answer ......................................................................
(Total 3 marks)
Q4.
(a)
Expand
w(w + 6)
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Answer ......................................................................
(2)
Page 2 of 28
(b)
Factorise fully
8y + 20
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Answer ......................................................................
(2)
(Total 4 marks)
Q5.
(a)
Expand and simplify
2(3x - 2) + 4(x + 5)
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Answer .................................................
(2)
(b)
Solve the equation
2(3x - 2) + 4(x + 5)
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Answer x = .................................................
(3)
(Total 5 marks)
Q6.
(a)
Expand and simplify
4(2x – 1) + 3(x + 6)
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Answer .................................................
(2)
(b)
Expand
x 2(4 – 2x)
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Answer .................................................
(2)
(Total 4 marks)
Page 3 of 28
Q7.
Show that
x(y + 6) – (xy + 4) ≡ 2(3x – 2)
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(Total 3 marks)
Q8.
(a)
Expand and simplify
3(2x + 4) + 2(x + 1)
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Answer ......................................................................
(3)
(b)
Factorise
x 2 − 11x
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Answer ......................................................................
(1)
(Total 4 marks)
Q9.
(a)
Multiply out
8(y + 3)
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Answer ......................................................................
(1)
(b)
Factorise
4x – x 2
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Answer ......................................................................
(1)
(Total 2 marks)
Page 4 of 28
Q10.
Expand and simplify
(2x − 3y)(4x − 5y)
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Answer ......................................................................
(Total 3 marks)
Q11.
(a)
Factorise
3x − 15
Answer ......................................................................
(1)
(b)
Multiply out
5(y + 4t − 2)
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Answer ......................................................................
(2)
(c)
Solve
3(w + 2) = 2w − 1
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w = ............................................................................
(3)
(Total 6 marks)
Q12.
(a)
Expand and simplify
(3x + 2)(2x + 5)
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Answer ......................................................................
(2)
Page 5 of 28
(b)
Simplify fully
(3x 2y 4)2
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Answer ......................................................................
(2)
(Total 4 marks)
Q13.
(a)
Multiply out and simplify
(x − 6)(x − 5)
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Answer .....................................................................
(2)
(b)
Simplify fully
2a2b3 × 4a5b6
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Answer .....................................................................
(2)
(Total 4 marks)
Q14.
(a)
Expand
m(m + 4)
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Answer ......................................................................
(2)
(b)
Factorise fully
12xy 2 − 6y
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Answer ......................................................................
(2)
(Total 4 marks)
Page 6 of 28
Q15.
(a)
(x + 6)2
Expand and simplify
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Answer ......................................................................
(2)
(b)
Expand and simplify
9w(3x − 4y ) − 5w(x + y )
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Answer ......................................................................
(4)
(Total 6 marks)
Q16.
(a)
Tom finds the value of
Sam finds the value of
(2n – 1)(n + 1)
(2n – 1)(n + 1)
when n = 1
when n = 2
Work out the difference between Tom’s value and Sam’s value.
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Answer .......................................................................
(3)
(b)
Expand and simplify
(2n – 1)(n + 1)
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Answer .......................................................................
(2)
(Total 5 marks)
Page 7 of 28
Q17.
(a)
w(w − 4)
Expand
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Answer ......................................................................
(2)
(b)
Factorise
8t + 24
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Answer ......................................................................
(1)
(c)
Expand and simplify
(y + 7)(y − 2)
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Answer ......................................................................
(2)
(Total 5 marks)
Q18.
Expand and simplify
(3x + y )(2x − 5y )
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Answer ......................................................................
(Total 3 marks)
Q19.
(a)
Factorise
7x − 21
Answer ......................................................................
(1)
(b)
Multiply out
4(y + 9)
Answer ......................................................................
(1)
(Total 2 marks)
Q20.
(a)
Factorise
5x 2 + 20x
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Answer .................................................
(1)
Page 8 of 28
(b)
Factorise
x 2 – 49
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Answer .................................................
(1)
(c)
Factorise fully
(3x + 4)2 – (2x + 1)2
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Answer .................................................
(3)
(Total 5 marks)
Q21.
(a)
Expand and simplify
(2x + 1)(x − 2)
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Answer .....................................................................
(3)
(b)
Factorise fully
3x 2 − 48y2
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Answer .....................................................................
(3)
(Total 6 marks)
Page 9 of 28
Q22.
(a)
(2x + 1)(3x − 4)
Expand and simplify
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Answer ......................................................................
(2)
(b)
Factorise
6x 2 − 23x − 4
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Answer ......................................................................
(2)
(Total 4 marks)
Q23.
Multiply out and simplify
(2p – 5q)(3p + q)
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Answer .................................................
(Total 3 marks)
Q24.
(a)
Show clearly that (3x + 1)2 ≡ 9x 2 + 6x + 1
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(1)
Page 10 of 28
(b)
Solve the simultaneous equations y = 3x + 1
y2 = 4x2 − x + 7
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Answer .....................................................................
(5)
(Total 6 marks)
Q25.
Here is an identity
(3x + c)(x + c) ≡ 3x 2 − dx + 16
c and d are integers.
Work out all possible pairs of values of c and d .
You must show your working.
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Answer ......................................................................
(Total 5 marks)
Page 11 of 28
Q26.
(a)
Show clearly that
(p + q)2 ≡ p 2 + 2pq + q2
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(1)
(b)
Hence, or otherwise, write the expression below in the form ax2 + bx + c
(2x + 3)2 + 2(2x + 3)(x – 1) + (x – 1)2
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Answer .................................................
(3)
(Total 4 marks)
Page 12 of 28
w2 + 6w
M1.
B1 for w2 or (+) 6w
Award B1 if further working seen after correct answer
B2
[2]
M2.
(a)
3x − 18
B1
(b)
5(y − 2)
B1
(c)
12w + 3 − 15w + 10
(12w + 3) − (15w − 10)
Allow one sign or arithmetic error for M1
M1
12w + 3 − 15w + 10
A1 if all correct
A1
− 3w + 13
ft their expansion if M awarded
Ignore any non-contradictory further work, such as solving an
equation, but do not award A1 if contradictory further work, such as
= 10w
A1ft
[5]
M3.
5x − 15 − 2x + 2
Attempt to expand both brackets to 4 terms with at least 3 correct
M1
5x − 15 − 2x + 2
A1 if fully correct
A1
3x − 13
ft on one error
A1ft
[3]
M4.
(a)
w2+ 6w
B1 for w2 or (+) 6w
Award B1 if further working seen after correct answer
B2
Page 13 of 28
(b)
4(2y + 5)
B1 for 2(4y + 10) or 8(y + 2.5)
Award B1 if further working seen after correct answer
B2
[4]
M5.
(a)
Allow one error
M1
Ignore fw that does not contradict, but do not
award A1 for fw such as = 26x
A1
(b)
Allow 1 error
ft Their answer for (a)
ie, Their (a) = 4x – 8
M1
–4
ft on one error only for A1
Errors can be in expansion (1 error)
Collecting terms to ax = b
Solving equation
A2 ft
[5]
M6.
(a)
8x – 4 + 3x + 18
Allow one error
M1
11x + 14
fw that does not contradict is not penalised but fw
such as = 25x do not award A1
A1
Page 14 of 28
(b)
4x 2 – 2x 3
B1 each term fw such as = 2x5 only give B1
4x2 – 2x2 = 2x2 is B1, 4x2– 2x2 = 6x4is BO
B2
[4]
6x − 4
M7.
B1
LHS = xy + 6x − xy − 4
Both brackets must be removed.
Must see xy and − xy
Allow +4 for B1
B1
Expanding LHS and simplifying and
stating
Strand (ii). For the Q mark this must be clearly shown and not
‘assumed’.
6x − 4 = 2(3x − 2)
or 2(3x − 2) = 6x − 4
or showing clearly that all terms cancel.
If + 4 seen in expansion and this is subsequently changed to −4 do
not allow the Q mark unless the error is recognised and
‘recovered’.
Q1
[3]
M8.
(a)
6x + 12 or 2x + 2
M1
6x + 12 + 2x + 2
A1
8x + 14 oe
ft from their 4 terms
A1ft
(b)
x (x − 11) or (x − 11) x
B1
[4]
M9.
(a)
8y + 24
or
24 + 8y
B1
Page 15 of 28
(b)
x (4 − x )
B1
[2]
8x 2 − 12xy − 10xy + 15y2
Allow one term error
M10.
M1
8x 2 − 12xy − 10xy + 15y2
A1
8x 2 − 22xy + 15y2
ft their four terms if M1 awarded
Do not ignore fw for final mark
A1 ft
[3]
M11.
(a)
3(x − 5)
B1
(b)
5y + 20t − 10
B1 for 2 correct terms.
Penalise any incorrect further working.
Eg
5y + 20t − 10 = 25yt − 10 is B1
5y + 20t − 1 = 25yt − 1 is B0 (error in expansion and incorrect
further work)
5y + 20t − 10 = 5(y + 4t − 2) given as answer is B1 as shows a
misunderstanding of expanding brackets.
B2
(c)
3w + 6 = 2w − 1
w+2=
w − M1
3w − 2w = −1 − 6
This mark is for rearranging their expansion correctly to get w
terms one side and number terms on the other.
w − w = − − 2 (oe)
M1
−7
ft on one error
A1ft
[6]
Page 16 of 28
M12.
(a)
6x 2 + 4x + 15x + 10
Allow one sign or arithmetic error. Must see 4 terms including term
in x 2, 2 terms in x and a constant term
M1
6x 2 + 19x + 10
NB Answer only
6x 2 + 19x + b implies M1
ax 2 + 19x + 10 implies M1
Do not award if incorrect further work
A1
(b)
9x 4y8
B1 for two of 9, x 4 or y8
B1 maximum for any use of × signs
B0 for any addition eg 9 + x 4 + y8
Deduct one mark for incorrect further work
B2
[4]
M13.
(a)
x 2 − 5x − 6x + 30
four terms, three correct with a term in x 2 or x 2 − 11x + k with k ≠ 0
M1
x 2 − 11x + 30
A1
(b)
8a7b9
B1 two correct from 8, a7 and b9
B1 correct answer with multiplication sign(s)
B2
[4]
M14.
(a)
m2 + 4m
B1 for one term correct
B2
(b)
6y (2xy − 1)
oe
B1 for 6(2x 2 − y ) or 3(4xy 2 − 2y )
or 2(6xy 2 − 3y ) or y (12xy − 6) or 3y (4xy − 2)
or 2y (6xy − 3) or 6y (? − ?) eg 6y (2xy − y )
B2
[4]
Page 17 of 28
M15.
x 2 + 6x + 6x + 36
(a)
Allow one error
M1
x 2 + 12x + 36
Do not ignore further working
A1
(b)
27wx − 36wy or − 5wx − 5wy
M1
27wx − 36wy − 5wx − 5wy
A1
22wx − 41wy or w(22x − 41y )
ft only if 3 of the 4 terms are correct
Do not ignore further working
A1ft
Correct symbolic notation for their simplified answer
Strand (i)
Must contain terms in wx and wy only
Q1
[6]
M16.
(a) (2 × 1 – 1) × (1 + 1)
or
(2 × 2 – 1) × (2 + 1)
or 2 × 12 + 1 – 1
or 2 × 22 + 2 – 1
M1
2 or 9
A1
(±) 7
A1
(b)
2n2 + n – 1
B1 2n2 + 2n – n – 1 any three out of four terms correct
B2
[5]
M17.
(a)
w2 − 4w
B1 for w2 or − 4w
B2
Page 18 of 28
(b)
8(t + 3)
Accept 4 (2t + 6) or 2(4t + 12)
B1
(c)
y2 − 2y + 7y − 14
Allow one error
Must see 4 terms
M1
y2 + 5y − 14
A1
[4]
6x 2 − 15xy + 2xy − 5y 2
3 terms correct
M18.
M1
6x 2 − 15xy + 2xy − 5y 2
A1
6x 2 − 13xy − 5y 2
ft from four terms
A1 ft
[3]
M19.
(a)
7(x − 3)
B1
(b)
4y + 36
B1
[2]
M20.
(a)
5x (x + 4)
B1
(b)
(x + 7)(x – 7)
B1
Page 19 of 28
(c)
M1 for expanding and collecting to general quad form,
allow one error but expansions must have x 2 term,
x term and constant term.
Allow misuse of minus.
eg 9x2 + 24x + 16 – 4x2 + 4x + 1
Difference of two squares
((3x + 4) – (2x + 1)) × ((3x + 4) + (2x + 1))
M1
5x 2 + 20x + 15
A1 for either (x + 3) or (5x + 5) if difference of 2 squares used.
A1
5 (x + 3)(x + 1)
Accept
(x + 3)(5x + 5) or (5x + 15)(x + 1)
A1
[5]
M21.
(a)
2 x 2 + x – 4x – 2
4 terms, allow one error but must have a term in x 2
M1
2x 2 + x – 4x – 2
A1
2x 2 – 3x – 2 oe
ft their 4 terms if M1 awarded
SC1 answer of
2x 2 – 5x – 2 or 2x 2 + 3x – 2 or
2x 2 – 3x + 2
without working worth at least M1
A1 fit
(b)
3(x 2 − 16y2)
M1
(3)(x + ay)(x + by)
where ab = – 16
M1
3(x − 4y)(x + 4y) oe
A1
Alternative method
(3x + ay)(x + by)
where ab = – 48
M1
Page 20 of 28
(3x + 12y)(x − 4y)
or
(3x −12y)(x + 4y)
M1
3(x − 4y)(x + 4y) oe
A1
[6]
M22.
(a)
6x 2 + 3x − 8x − 4
Must have 4 terms shown or implied, including a quadratic term,
two linear terms and a constant term. Could be in a grid from box
method
Allow one sign or arithmetic error for M1
M1
6x 2 − 5x − 4
kx2 − 5x − 4 or 6x2 − 5x − k both imply M1
A1
(b)
(ax ± c)(bx ± d )
ab = 6, cd = 4 or −4
6x (x − 4) + (1)(x − 4)
x(6x + 1) − 4(6x + 1)
M1
(6x + 1)(x − 4)
Ignore any subsequent attempt to solve once the correct
factorisation seen
A1
[4]
M23.
6p2 + 2pq – 15pq – 5q2
For 3 correct terms
M1
6p2 + 2pq – 15pq – 5q2
Fully correct
A1
6p2 – 13pq – 5q2
From 4 terms
Do not ignore fw
B1 ft
[3]
Page 21 of 28
M24.
(a)
(3x + 1)2 = 9x 2 + 3x + 3x + 1
B1
(b)
9x 2 + 3x + 3x + 1 = 4x 2 − x + 7 or 9x 2 + 6x + 1 = 4x 2 − x + 7
oe
B1
5x 2 + 7x − 6 = 0
ft their expansion of (3x + 1)2 with all terms correctly
collected on one side of the equation
M1
(5x − 3)(x + 2) (= 0) or (5x + a )(x + b ) (= 0)
ab = ±6 or 5b + a = ±7 ft their quadratic
or quadratic formula allowing one substitution error
M1
x = 0.6 and x = −2 or x = 0.6 and y = 2.8
oe
A1
y = 2.8 and y = −5 or x = −2 and y = −5
oe
A1
[6]
M25.
c2 = 16 or c = 4 or c = −4
M1
3x 2+ 3cx + cx + c2 (= 3x 2 − dx + 16)
3x 2 + 12x + 4x + 16 or 3x 2 − 12x − 4x + 16 oe
M1
c = 4 and c = −4 or 4c = −d
or 16 = −d or −16 = −d
oe
M1
c = 4 and d = −16
or
c = −4 and d = 16
One pair of answers or
all four answers seen but not paired
A1
Page 22 of 28
c = 4 and d = −16
and
c = −4 and d = 16
Both pairs of answers must be correctly paired
SC3 for one correct pair or both correct pairs or all four answers
seen but not paired from no working
A1
[5]
M26.
(a)
Convincing algebra
Must see
or
box method and
B1
(b)
Allow one sign or coefficient error
For middle term accept
or
M1
A1
ft if M1 awarded and no further errors
A1 ft
[4]
Page 23 of 28
E1.
E2.
This question was accessible to most students. A significant number of students gave 2w
for w × w. Some students, having obtained two terms went on to try and combine them, giving,
2
for example, 6w . Some used incorrect terminology such as w6.
Foundation Tier
Students at this tier found the algebra challenging. Factorising in part (b) was beyond the
capability of the vast majority. In part (c) there were a small number of promising solutions seen.
In these, often three of the four terms were correct and some went on to collect their terms
correctly. Students’ presentation was sometimes unclear, with some working on the brackets
separately and usually omitting the subtraction sign.
Higher Tier
Parts (a) and (b) were usually answered correctly. In part (c), expanding –5 × –2 as –10 was the
most common error. Many errors with minus signs were seen.
E3.
Foundation Tier
The method of multiplying out brackets and simplifying was generally well known. Many students
were able to correctly expand three of the four terms and progress to correctly collect these
terms. Sometimes working was unclear, particularly when students expanded the two brackets
separately and omitted the subtraction sign when combining them.
Higher Tier
The majority of students knew to expand and collect terms, but many arithmetic errors were
seen. The main one of these was –2 × –1 = –2.
E4.
This question was well answered. In part (a) there were a few instances of w6, instead of
6w. In part (b), some only removed a common factor of 2 and some used a common factor of 8.
E5.
Part (a) was well done with just about all candidates picking up at least 1 mark and the vast
majority picking up 2. Common errors were, for example, simplifying 6x – 4 +4x + 20 to
10x – 24 or making an error in expanding, such as 5x – 4 +4x + 20. (One error was allowed for
the method mark). Some candidates, having obtained 10x + 16, then divided by 2 to give an
answer of 5x + 8 and so were penalised for incorrect further working.
Part (b) was interesting to mark, as follow through was allowed from the answer in part (a). The
vast majority of candidates did use the answer to (a) = 4x – 8 but then a variety of errors
ensued, with about a third of the candidates making one of the following errors: expanding the
right-hand side to 4x – 2 or 4x – 6; rearranging the terms incorrectly to the form ax = b, for
example 6x = 16; solving the equation incorrectly, for example 6x = 10, x = 0.6. If only one error
was made, 2 out of 3 marks were possible. Two errors earned 1 mark out of the 3 possible.
Page 24 of 28
E6.
Intermediate Tier
Most candidates scored 1 mark in each part. Many gave expressions involving 5x or 22 and
8x – 4 + 3x + 9 was common in part (a). Most got 4x 2 in part (b) but usually with 2x 2 and some
tried to factorise as (x – 2)(x – 2).
Higher Tier
This question was usually done well. Errors were mainly due to carelessness with numeracy,
powers or minus signs.
E7.
This is a Quality of Written Communication question, which means that clear and accurate
algebra needs to be seen. The most common answer was to expand the left hand side as xy +
6x − xy + 4, expand the right hand side as 6x − 4 and then state that 6x + 4 = 6x − 4. Rarely
was the obvious inconsistency corrected, although many ‘plus’ signs were overwritten by a
‘minus’ sign. Students should be discouraged from overwriting.
E8.
This question was well answered by the majority of students. Most gave a fully correct
solution in part (a) or missed the last multiplication and obtained 6x + 12 + 2x + 1 = 8x + 13.
Some students, after obtaining the correct answer, went on to do further incorrect work,
simplifying to 4x + 7 or solving 8x + 14 = 0.
E9.
Both parts provided an excellent start to the paper for most students.
E10.
This question was also a good discriminator. The most common errors were incorrect
indices for 8x 2 or 15y2 or making 15y2 negative. Some students added the coefficients resulting in
6x − 7xy − 7xy − 8y or 6x − 8y.
E11.
All parts of this question were well answered. Rearrangement errors were quite common in
part (c).
E12.
Both parts of this question were poorly answered. In part (a), the main errors were 2x × 3x
= 6x leading to an answer of 25x + 10, or simply adding the two brackets giving an answer of 5x
+ 7. In part (b), errors were made with the number term, 3 and 6 being common and in
multiplying the powers with y 6 being common.
E13.
The majority of students gave a fully correct solution. Poor arithmetic led to common errors,
for example, incorrectly multiplying −6 and −5 to achieve −30, or incorrectly collecting terms
after a correct expansion for −6x and −5x with +11x , + x or − x as the term in x .
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E14.
Part (a) was well answered. Most students attempted part (b) but with mixed success.
Common errors were usually due to only partial factorisation. A few tried to give their answer as
a product of two brackets, treating it as a standard quadratic.
E15.
The standard of algebra on both parts was often poor. In part (a), x 2 + 36 was a common
incorrect answer. Part (b) was a very good discriminator, although the number of students with
fully correct answers was low. Common errors included sign errors (particularly with the last
term), arithmetic errors and conceptual errors such as combining all the letters into a single
term for their final answer.
E16.
On the whole, part (a), was well answered with over sixty percent gaining full marks. Some
candidates substituted correctly but then added the brackets instead of multiplying.
In part (b) about half the candidates managed this expansion successfully but over thirty percent
showed no appreciation of what to do to complete this routine expansion. A significant number of
those candidates who followed a correct method made silly mistakes when collecting the terms.
E17.
Parts (a) and (b) of this question were answered very well. In part (c), using ‘FOIL’ to
expand the brackets was more popular than the grid method. Some candidates went straight for
the three-term answer, although this can be a risky strategy. Even in the grid method there were
2
many answers of 5 instead of ‒14, making y + 5y + 5 a very common answer. Other common
2
2
2
answers were, y + 5, y ‒ 14, y + 9y ‒ 14 and even y + 7 + y ‒ 2 = 2y + 5. A few candidates
2
worked out the correct answer but then went on to give a final answer, such as 6y ‒ 14.
E18.
E19.
E20.
Approximately half of all students did not score marks on this question. Common errors
amongst those who made some progress were to write 5x 2, + 15xy or − 5y . These errors were
more frequent when a grid method was used.
Both parts were well answered.
Parts (a) and (b) were usually correct, with (x – 7)2 a common error in (b). Part (c) was
done correctly by less than half the candidates. Hardly any candidates used the difference of 2
squares to answer this. The vast majority tried to expand, subtract and then factorise. Although
the ‘invisible bracket’ for the expansion of the second bracket
(ie, 9x 2 + 24x + 16 – 4x 2 + 4x + 1) was tolerated for the method mark, unless it was recovered
no further marks could be scored on this question. Pleasingly many candidates did recover this
and went on to give a complete answer. Others got to the stage of 5x 2 + 20x + 15 and either
cancelled 5 or left the answer as 5(x 2 + 4x + 3).
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E21.
E22.
The expansion in part (a) was quite well answered although there were errors when
collecting terms. Part (b) was poorly answered with very few fully correct answers. Most
students only took out the factor 3.
Part (a) was quite well answered. Many students did not have a method for expanding two
brackets. Of those who did the x 2 term was often omitted. The other main errors were with signs
and / or arithmetic. Part (b) was not well answered. Many students did not have a method for
factorising a quadratic. Of those who had some idea to find x terms with a product of 6x 2, the
constant terms were often added to give −4.
E23.
Responses to this question were usually poor with only the very best candidates scoring.
Very few candidates made any reasonable attempt to multiply out brackets.
E24.
In part (a), many students correctly showed the expansion of the brackets, with 3x + 3x
being the key terms required. Writing 2 × 3x was not accepted as clear evidence of a correct
method. A correctly completed grid method was accepted. Very few students used part (a) to
help them in part (b), and most restarted by expanding (3x + 1)2 again. Hence, many errors
were seen in the first step of expanding the brackets, which would have been avoided if part (a)
had been used. Poor algebraic skills when rearranging terms and factorising the resulting
quadratic equation were common errors. A small minority correctly calculated both solutions.
The y values were sometimes omitted from otherwise completely correct responses.
E25.
Students found this question challenging. Those who attempted the question usually made
some progress correctly expanding the pair of brackets on the left hand side of the identity, and
some also realised that c2 = 16, hence c = 4 with the negative solution often being omitted.
Some students did see that the connection between c and d was 4c = ‒d but very few students
scored more than 2 marks for the question.
E26.
Part (a) was a ‘lead in’ to part (b) and was well done by the majority of candidates. Basically
the terms pq and qp needed to be seen. Despite the lead in and the ‘hence’, all candidates opted
for the ‘or otherwise’ approach which was to expand the brackets. Many candidates did this
successfully and one minor error was allowed which was usually followed through thus giving 2
marks out of 3. The most common error was a failure to deal with the coefficient of 2 in the
second term. This was usually multiplied into both brackets or on many occasions candidates
attempted to expand the brackets and multiply by 2 at the same time with an inevitable 2 errors
which meant zero marks. Other errors were (2x + 3)2 = 2x 2 + 9, which suggests that candidates
did not link the two parts of the question. ‘Hence’ means that in the view of the examiner using a
previous result will be the easiest approach. The ‘or otherwise’ approach will always take longer
and have more opportunities for error.
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