Higher June 2022 Paper 2 Q14
14
(a) Expand and simplify \((5 - x)(2x + 3)(x + 4)\)
Show your working clearly. (3)
Show your working clearly. (3)
(b) Make \(c\) the subject of \(g = \dfrac{c + 3}{4 + c} - 7\) (4)
| Scheme | Marks |
|---|---|
| \((5 - x)(2x + 3) = 10x + 15 - 2x^2 - 3x\;(= -2x^2 + 7x + 15)\) or \((5 - x)(x + 4) = 5x + 20 - x^2 - 4x\;(= -x^2 + x + 20)\) or \((2x + 3)(x + 4) = 2x^2 + 8x + 3x + 12\;(= 2x^2 + 11x + 12)\) | M1 |
| \((-2x^2 + 7x + 15)(x + 4) = -2x^3 - 8x^2 + 7x^2 + 28x + 15x + 60\) or \((-x^2 + x + 20)(2x + 3) = -2x^3 - 3x^2 + 2x^2 + 3x + 40x + 60\) or \((2x^2 + 11x + 12)(5 - x) = 10x^2 - 2x^3 + 55x - 11x^2 + 60 - 12x\) | M1 |
| Working required Answer: \(-2x^3 - x^2 + 43x + 60\) | A1 |
| (3) |
Notes
M1: multiplying 2 factors only but do not award if they multiply eg \((5 - x)(2x + 3)\) and \((5 - x)(x + 4)\) as their method
allow one error
M1: (dep)ft for expanding by the third factor, allow one further error
A1: Dep on M1
ALTERNATIVE
| Scheme | Marks |
|---|---|
| \(10x^2 + 15x + 40x + 60 - 2x^3 - 3x^2 - 8x^2 - 12x\) | M2 |
| Working required Answer: \(-2x^3 - x^2 + 43x + 60\) | A1 |
Notes
M2: for a complete expansion with 8 terms present, at least 4 of which must be correct
(M1 for 4 correct terms from any number of terms)
| Scheme | Marks |
|---|---|
\(g + 7 = \dfrac{c + 3}{4 + c}\) or \(g(4 + c) = c + 3 - 7(4 + c)\) or \(g = \dfrac{c + 3}{4 + c} - \dfrac{7(4 + c)}{4 + c}\left(= \dfrac{c + 3 - 28 - 7c}{4 + c}\right)\) | M1 |
| eg \(4g + gc + 28 + 7c = c + 3\) or \(4g + gc = c + 3 - 28 - 7c\) oe | M1 |
| eg \(gc + 7c - c = 3 - 28 - 4g\) or \(28 - 3 + 4g = c - 7c - gc\) | M1ft |
| \(c = \dfrac{-(4g + 25)}{g + 6}\) | A1 |
| (4) | |
| (7 marks) |
Notes
M1: Adding 7 to both sides as a first step or removing fraction correctly
M1: removing fraction and expanding all brackets in an equation with no more than one error
M1ft: ft dep on previous M1 - terms in \(c\) on one side and other terms on the other side in an equation
A1: oe eg \(c = \dfrac{25 + 4g}{-6 - g}\) or \(c = \dfrac{3 - 28 - 4g}{g + 7 - 1}\) oe
[if c = is missing allow full marks if seen in working otherwise 3 marks]
(SCB2 for an answer of \(c = \dfrac{-4 - 4g}{g - 1}\) oe or \(c = \dfrac{31 - 4g}{g - 8}\) oe
SCB1 in working for \(4g + cg = c + 3 - 7\) oe or \(4g + cg - 28 - 7c = c + 3\) oe