Higher June 2021 Paper 1 Q13
13 Expand and simplify \(4x(3x + 1)(2x - 3)\)
Show your working clearly.
(3)
| Scheme | Marks |
|---|---|
| \(4x(3x + 1) = 12x^2 + 4x\) or \(4x(2x - 3) = 8x^2 - 12x\) or \((3x + 1)(2x - 3) = 6x^2 - 9x + 2x - 3\ (= 6x^2 - 7x - 3)\) | M1 |
| \((12x^2 + 4x)(2x - 3) = 24x^3 - 36x^2 + 8x^2 - 12x\) oe \((8x^2 - 12x)(3x + 1) = 24x^3 + 8x^2 - 36x^2 - 12x\) oe \(4x(6x^2 - 7x - 3) =\) eg \(24x^3 - 28x^2 \ldots\) oe | M1 |
| Working required Answer: \(24x^3 - 28x^2 - 12x\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for expanding two of the three factors, allow one error
M1: (dep)ft for expanding by the third factor, allow one error
(some may do the expansion in one stage and will get to \(24x^3 - 36x^2 + 8x^2 - 12x\) without firstly expanding two factors)
A1: dep on M1
isw correct factorisation eg
\(4(6x^3 - 7x^2 - 3x)\)
\(x(24x^2 - 28x - 12)\)
\(4x(6x^2 - 7x - 3)\)
do not isw incorrect simplification eg \(24x^3 - 28x^2 - 12x = 6x^3 - 7x^2 - 3x\) gets M2A0