Expanding Brackets: WJEC GCSE Higher November 2022 Unit 1 Q4

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Higher November 2022 Unit 1 Q4

WJECHigherCurrent spec6 marksExpanding Brackets Indices Simplifying Basic Expressions

4. Simplify the following expressions.

(a) \(2p^3q \times 3p^4q^7\) [2]

(b) \(7a(a + 5) - 2(3a^2 + 6a - 7)\) [4]

Mark scheme (a)

Scheme	Marks	Comments
\(6p^7q^8\)	B2	Mark final answer Award B1 for one of the following: \(6 \times p^7 \times q^8\) \(6p^7 \times q^8\) \(6{\times}p^7q^8\) \(6p^7q^{\cdots\cdots}\) \(6p^{\cdots}q^8\) \(kp^7q^8\) (\(k \neq 0\) or 6) Sight of \(6p^7\) AND \(q^8\) in an expression (e.g. \(6p^7 + q^8\)).

Mark scheme (b)

Scheme	Marks	Comments
\(7a^2 + 35a - 6a^2 - 12a + 14\)	B2	Award B1 for one of the following: sight of \(7a^2 + 35a\). sight of \(-\,6a^2 - 12a + 14\) (brackets must be removed) Note: If \(7a^2 + 35a \;-\; 6a^2 + 12a - 14 = a^2 + 23a + 14\) is seen, then award B2 B2 (brackets implied).
\(= a^2 + 23a + 14\)	B2	FT for B2 if at least two \(a^2\) terms AND at least two \(a\) terms to be simplified. FT for B1 if at least two \(a^2\) terms OR at least two \(a\) terms to be simplified. Award B2 for \(1a^2 + 23a + 14\) If B2 not awarded, award B1 for one of the following: correct collection of ‘\(a^2\) terms’ \((1)a^2\) correct collection of ‘\(a\) terms’ \((+23a)\). This 2^nd B2 (or B1) is for their final answer. A correct answer must come from correct workings seen, however \(7a^2 + 35a \;-\; 6a^2 + 12a - 14 = a^2 + 23a + 14\) is awarded B2 B2 (brackets implied). Mark final answer Penalise −1 from the final B1 or B2 mark for any one of the following: incorrect subsequent working any attempt to equate their expression to zero (and attempting to solve) incorrectly factorising Note (sign error): Award B1B2 for \(7a^2 + 35a - 6a^2 + 12a + 14 = a^2 + 47a + 14\) \(7a^2 + 35a - 6a^2 + 12a - 14 = a^2 + 47a - 14\) \(7a^2 + 35a - 6a^2 - 12a - 14 = a^2 + 23a - 14\)

Scheme

Marks

Comments

\(7a^2 + 35a - 6a^2 - 12a + 14\)

Award B1 for one of the following:

sight of \(7a^2 + 35a\).
sight of \(-\,6a^2 - 12a + 14\) (brackets must be removed)

Note:
If \(7a^2 + 35a \;-\; 6a^2 + 12a - 14 = a^2 + 23a + 14\) is seen, then award B2 B2 (brackets implied).

\(= a^2 + 23a + 14\)

FT for B2 if at least two \(a^2\) terms AND at least two \(a\) terms to be simplified.
FT for B1 if at least two \(a^2\) terms OR at least two \(a\) terms to be simplified.

Award B2 for \(1a^2 + 23a + 14\)

If B2 not awarded, award B1 for one of the following:

correct collection of ‘\(a^2\) terms’ \((1)a^2\)
correct collection of ‘\(a\) terms’ \((+23a)\).

This 2^nd B2 (or B1) is for their final answer.
A correct answer must come from correct workings seen, however
\(7a^2 + 35a \;-\; 6a^2 + 12a - 14 = a^2 + 23a + 14\)
is awarded B2 B2 (brackets implied).

Mark final answer
Penalise −1 from the final B1 or B2 mark for any one of the following:

incorrect subsequent working
any attempt to equate their expression to zero (and attempting to solve)
incorrectly factorising

Note (sign error):
Award B1B2 for
\(7a^2 + 35a - 6a^2 + 12a + 14 = a^2 + 47a + 14\)
\(7a^2 + 35a - 6a^2 + 12a - 14 = a^2 + 47a - 14\)
\(7a^2 + 35a - 6a^2 - 12a - 14 = a^2 + 23a - 14\)

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