Foundation June 2022 Paper 2 Q21

(a) \[\boxed{\;(x + 4)(x + 3) = x^2 + 7x + 12\;}\]

Darcy says that the statement in the box is an equation.
Ellis says that the statement in the box is an identity.
One of them is correct.

Explain which one of Darcy or Ellis is correct. [2]

(b) Solve by factorising.

\(x^2 + 4x - 12 = 0\) [3]

Mark scheme (a)

Answer	Marks	Part marks and guidance
\(x^2 + 4x + 3x + 12\) [\(= x^2 + 7x + 12\)]	M1		May be seen in the work space or within their statement and it might be seen in e.g. a table where the addition can be implied
Ellis with a correct statement about definition of an identity or of an equation or Ellis with correct supporting work and statement	B1		e.g. an identity is true for all values of \(x\) or e.g. this cannot be solved to get a value of \(x\) or e.g. left-hand side is the same as the right-hand side Accept “neither, because if it was an identity then it would have ≡” See Appendix
		Alternative Method M1 for shows LHS = RHS for at least three values of \(x\) B1 dep for Ellis with their correct statement e.g. a quadratic equation does not have three solutions [so by elimination it is an identity]

Exemplar responses for Question 21a

	Response for the B marks	Mark
1	Ellis – the expression before and after the equal sign are the same (‘same’ implies identical in form)	1
2	Ellis – \((x + 4)(x + 3)\) is the same as \(x^2 + 7x + 12\)	1
3	Ellis – the statement in the box doesn’t gives a value, an equation would be to find a specific number	1
4	Identity – \((x+4)(x+3)\) has the same answer expanded as \(x^2 + 7x + 12\) (condone identity instead of Ellis)	1
5	Ellis – they are written differently but are equal to the same (this is ok to score as they have clarified that the 2 sides are the same but in different forms)	1 BOD
6	Ellis – \((x + 4)(x + 3)\) is equal \(x^2 + 7x + 12\) when expanded	0
7	Ellis there is nothing to work out (not enough)	0
8	Ellis – it is identifying how each formula is equal to each other (equal to each other is not enough)	0
9	Ellis – you’re not asked to work anything out, it gives you the answer (they say you’re not asked to work it out rather than saying it cannot be solved)	0

Mark scheme (b)

Answer	Marks	Part marks and guidance
\((x + 6)(x - 2)\)	M2	M1 for \((x + a)(x + b)\) where \(a + b = 4\) or \(ab = -12\) or for \(x + 6\) [= 0] and \(x - 2\) [= 0]	For M2 or M1 condone omission of final bracket e.g. \((x + 1)(x + 3)\) as \(a + b = 4\) or e.g. \((x - 3)(x + 4)\) as \(ab = -12\)
\(-6\) and 2 final answer	B1FT	strict FT for correct solutions from their quadratic factors If 0 scored SC1 for answer \(\pm 2\) and \(\pm 6\)	e.g. FT \(x = -1\) and \(x = -3\) from \((x + 1)(x + 3)\)