Higher November 2020 Unit 1 Q6
6.
(a) Factorise \(x^2 - 7x + 12\), and hence solve \(x^2 - 7x + 12 = 0\). [3]
(b) Expand and simplify \((5x - 2)^2\). [2]
| Scheme | Marks | Comments |
|---|---|---|
| \((x - 4)(x - 3)\) | B2 | B1 for \((x \ldots 4)(x \ldots 3)\). Ignore ‘= 0’. |
| \((x =)\ 4\) AND \((x =)\ 3\) | B1 | Strict FT from their brackets. Allow the following. B2 for \(x - 4\ (=0)\) AND \(x - 3\ (=0)\) (B1) \((x =)\ 4\) AND \((x =)\ 3\) (B1) B1 for \(x + 4\ (=0)\) AND \(x + 3\ (=0)\) (B0) \((x =)\ {-4}\) AND \((x =)\ {-3}\) (B1) FT B1 if only \((x =)\ 4\) AND \((x =)\ 3\) seen. (B1) |
| Scheme | Marks | Comments |
|---|---|---|
| \(25x^2 - 20x + 4\) | B2 | Otherwise B1 for sight of \(25x^2 \pm kx + 4\) (allow \(k = 0\)) B1 for sight of \(25x^2 - 20x - 4\) Mark final answer. |