Foundation November 2018 Paper 2 Q15
15
(a) Multiply out.\[(3x - 2y)(x + y)\]
Give your answer in its simplest form. [3]
(b) \(3(2x + d) + c(x + 5) = 10x + 17\)
Work out the value of \(c\) and the value of \(d\). [5]
(c) Solve by factorising.\[x^2 - 7x + 10 = 0\]
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(3x^2 + xy - 2y^2\) final answer | 3 | M2 for \(3x^2 - 2xy + 3xy - 2y^2\) oe or M1 for two correct terms | Accept e.g. 3yx May be seen in a table for M1 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(c = 4\) \(d = {}^{-}1\) | 5 | M4 for \(3d + 20 = 17\) oe or M3 for \(6 + c = 10\) or \(3d + 5c = 17\) or \(6x + cx = 10x\) or M2 for \(6x + 3d + cx + 5c\) oe or M1 for \(6x + 3d\) or \(cx + 5c\) OR B3 for \(c = 4\) and B2 for \(d = {}^{-}1\) | Accept e.g. c5 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2, 5 nfww | 3 | M2 for \((x - 2)\) and \((x - 5)\) or M1 for \((x + a)\) and \((x + b)\) where \(ab = 10\) or \(a + b = {}^{-}7\) B1 ft their quadratic factors If 0 scored SC1 for answer \(\pm 2\) and \(\pm 5\) | |