simplifying - expanding brackets - rawmaths · · 2015-03-09expand and simplify€€€€5(x...

Simplifying - Expanding brackets

113 minutes

106 marks

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) Expand and simplify 3(4w + 1) − 5(3w − 2)

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Answer ...................................................................... (3)

(Total 5 marks)

Q3. Expand and simplify 5(x − 3) − 2(x − 1)

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Q4. (a) Expand w(w + 6)

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Answer ...................................................................... (2)

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(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 x2(4 – 2x)

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Answer ................................................. (2)

(Total 4 marks)

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Q7. Show that x(y + 6) – (xy + 4) ≡ 2(3x – 2)

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Q8. (a) Expand and simplify 3(2x + 4) + 2(x + 1)

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Answer ...................................................................... (3)

(b) Factorise x2 − 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 – x2

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Answer ...................................................................... (1)

(Total 2 marks)

of 28

Q10. Expand and simplify (2x − 3y)(4x − 5y)

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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)

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(b) Simplify fully (3x2y4)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 12xy2 − 6y

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Answer ...................................................................... (2)

(Total 4 marks)

of 28

Q15. (a) Expand and simplify (x + 6)2

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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 (2n – 1)(n + 1) when n = 1

Sam finds the value of (2n – 1)(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)

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Q17. (a) Expand w(w − 4)

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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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Q19. (a) Factorise 7x − 21

Answer ...................................................................... (1)

(b) Multiply out 4(y + 9)

Answer ...................................................................... (1)

(Total 2 marks)

Q20. (a) Factorise 5x2 + 20x

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Answer ................................................. (1)

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(b) Factorise x2 – 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 3x2 − 48y2

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Answer ..................................................................... (3)

(Total 6 marks)

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Q22. (a) Expand and simplify (2x + 1)(3x − 4)

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Answer ...................................................................... (2)

(b) Factorise 6x2 − 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 ≡ 9x2 + 6x + 1

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(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) ≡ 3x2 − 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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Q26. (a) Show clearly that (p + q)2 ≡ p 2 + 2pq + q2

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(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)

of 28

M1. w2 + 6w

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


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(b) 4(2y + 5)

B1 for 2(4y + 10) or 8(y + 2.5)


[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

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(b) 4x2 – 2x3

B1 each term fw such as = 2x5 only give B1

4x2 – 2x2 = 2x2 is B1, 4x2– 2x2 = 6x4is BO

B2 [4]

M7. 6x − 4 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

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(b) x (4 − x) B1

[2]

M10. 8x2 − 12xy − 10xy + 15y2

Allow one term error M1

8x2 − 12xy − 10xy + 15y2

A1

8x2 − 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]

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M12. (a) 6x2 + 4x + 15x + 10

Allow one sign or arithmetic error. Must see 4 terms including term

in x2, 2 terms in x and a constant term M1

6x2 + 19x + 10

NB Answer only

6x2 + 19x + b implies M1

ax2 + 19x + 10 implies M1

Do not award if incorrect further work A1

(b) 9x4y8

B1 for two of 9, x4 or y8

B1 maximum for any use of × signs

B0 for any addition eg 9 + x4 + y8

Deduct one mark for incorrect further work B2

[4]

M13. (a) x2 − 5x − 6x + 30

four terms, three correct with a term in x2 or x2 − 11x + k with k ≠ 0

M1

x2 − 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(2x2 − y) or 3(4xy2 − 2y)

or 2(6xy2 − 3y) or y(12xy − 6) or 3y(4xy − 2)

or 2y(6xy − 3) or 6y(? − ?) eg 6y(2xy − y) B2

[4]

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M15. (a) x 2 + 6x + 6x + 36

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

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(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]

M18. 6x2 − 15xy + 2xy − 5y2

3 terms correct M1

6x2 − 15xy + 2xy − 5y2

A1

6x2 − 13xy − 5y2

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

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(c) M1 for expanding and collecting to general quad form,

allow one error but expansions must have x2 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

5x2 + 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) 2x2 + x – 4x – 2

4 terms, allow one error but must have a term in x2

M1

2x2 + x – 4x – 2

A1

2x2 – 3x – 2 oe

ft their 4 terms if M1 awarded SC1 answer of

2x2 – 5x – 2 or 2x2 + 3x – 2 or

2x2 – 3x + 2

without working worth at least M1 A1 fit

(b) 3(x2 − 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

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(3x + 12y)(x − 4y)

or

(3x −12y)(x + 4y) M1

3(x − 4y)(x + 4y) oe A1

[6]

M22. (a) 6x2 + 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

6x2 − 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]

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M24. (a) (3x + 1)2 = 9x2 + 3x + 3x + 1

B1

(b) 9x2 + 3x + 3x + 1 = 4x2 − x + 7 or 9x2 + 6x + 1 = 4x2 − x + 7

oe B1

5x2 + 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

3x2+ 3cx + cx + c2 (= 3x2 − dx + 16)

3x2 + 12x + 4x + 16 or 3x2 − 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

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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]

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E1. 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,

for example, 6w . Some used incorrect terminology such as w6.

E2. 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.

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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 4x2 in part (b) but usually with 2x2 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 8x2 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 y6 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), x2 + 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

many answers of 5 instead of ‒14, making y + 5y + 5 a very common answer. Other common

answers were, y + 5, y ‒ 14, y + 9y ‒ 14 and even y + 7 + y ‒ 2 = 2y + 5. A few candidates

worked out the correct answer but then went on to give a final answer, such as 6y ‒ 14.

E18. Approximately half of all students did not score marks on this question. Common errors

amongst those who made some progress were to write 5x2, + 15xy or − 5y. These errors were more frequent when a grid method was used.

E19. Both parts were well answered.

E20. 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, 9x2 + 24x + 16 – 4x2 + 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 5x2 + 20x + 15 and either cancelled 5 or left the answer as 5(x2 + 4x + 3).

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2 2 2

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E21. 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.

E22. Part (a) was quite well answered. Many students did not have a method for expanding two

brackets. Of those who did the x2 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 6x2, 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 = 2x2 + 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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