Higher June 2025 Paper 1R Q8
8
(a) Simplify \(a^6 \times a^{10}\) (1)
(b) Simplify \(c^{30} \div c^{12}\) (1)
(c)
(i) Factorise \(y^2 - 10y + 21\) (2)
(ii) Hence, solve \(y^2 - 10y + 21 = 0\) (1)
| Scheme | Marks |
|---|---|
| \(a^{16}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(c^{18}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| (i) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \((y - 3)(y - 7)\) | A1 |
| (2) | |
| (ii) Answer: 3, 7 | B1 |
| (1) | |
| (5 marks) |
Notes
M1: for \((y \pm 3)(y \pm 7)\)
or for \((y \pm a)(y \pm b)\) with \(ab = 21\) or \(a + b = -10\)
A1: for correct factors
B1: ft dep on factorising in the form \((y \pm p)(y \pm q)\)