June 2019 Paper 2 Q25
25 \(\mathrm{f}(x) = x^3 - 10x - c \qquad\) where \(c\) is a positive integer.
\((x + c)\) is a factor of \(\mathrm{f}(x)\).
Use the factor theorem to work out the value of \(c\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((-c)^3 - 10(-c) - c\ (= 0)\) or \(-c^3 + 10c - c\ (= 0)\) or \(-c^3 + 9c\ (= 0)\) | M1 | oe |
| \(c(9 - c^2)\ (= 0)\) or \(c(3 + c)(3 - c)\ (= 0)\) or \(c^2 = 9\) | M1dep | oe factorised expression or quadratic equation |
| 3 with no other value(s) | A1 | SC2 answer 3 with one or both of \(-3\) and 0 and no other value |
| Alternative method 2 | ||
| \((x + c)(x^2 - cx - 1)\) | M1 | |
| \(-1 - c^2 = -10\) | M1dep | oe quadratic equation |
| 3 with no other value(s) | A1 | SC2 answer 3 with one or both of \(-3\) and 0 and no other value |
Additional guidance
| \((-3)^3 - 10(-3) - 3 = 0\) and Answer 3 (no part marks) | M2A1 |
| \((-3)^3 - 10(-3) - -3 = 0\) and Answer 3 | Zero |
| \(3^3 - 10(3) - -3 = 0\) and Answer 3 | Zero |
| Answer 3 with no incorrect working | M2A1 |
| Allow recovery of missing brackets |