Foundation June 2019 Paper 3 Q27
27 Solve by factorising.
\[x^2 + 3x - 10 = 0\][3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((x + 5)(x - 2)\) | M2 | or M1 for \((x \pm a)(x \pm b)\) where \((a + b) = 3\) or \((ab) = {}^{-}10\) | Eg \((x + 1)(x + 2)\) giving \(x^2 + 3x + 2\) or \((x - 1)(x + 10)\) giving \(x^2 + 9x - 10\) Eg FT \(x = -1\) and \(-2\) FT \(x = 1\) and \(-10\) |
| \({}^{-}5\) and 2 final answer | B1FT | for correct solutions from their quadratic factors If 0 scored SC1 for -5 and 2 as answers | |
Notes
(corrected from the printed mark scheme: the guidance printed \((x - 1)(x + 10)\) giving \(x^2 - 9x - 10\); it gives \(x^2 + 9x - 10\))