Higher June 2022 Paper 3 Q14
14
(a) Factorise fully \(4p^2 - 36\) (2)
(b) Show that \((m + 4)(2m - 5)(3m + 1)\) can be written in the form \(am^3 + bm^2 + cm + d\) where \(a\), \(b\), \(c\) and \(d\) are integers. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(4(p - 3)(p + 3)\) | M1 | for \(4(p^2 - 9)\) or partial factorisation which includes the product of 2 linear factors eg \((4p - 12)(p + 3)\) or \((p - 3)(4p + 12)\) or \((2p - 6)(2p + 6)\) or \(2(2p - 6)(p + 3)\) or \(2(2p + 6)(p - 3)\) or \(2(p - 3)2(p + 3)\) |
| A1 | for \(4(p - 3)(p + 3)\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(6m^3 + 11m^2 - 57m - 20\) | M1 | for a method to find the product of two linear expressions, 3 correct terms out of 4 terms, eg \(6m^2 + 2m - 15m - 5 = 6m^2 - 13m - 5\) or \(2m^2 + 8m - 5m - 20 = 2m^2 + 3m - 20\) or \(3m^2 + 12m + m + 4 = 3m^2 + 13m + 4\) |
| M1 | for a complete method to obtain all terms, at least half of which are correct (ft their first product), eg \(6m^3 + 2m^2 - 15m^2 + 24m^2 + 8m - 60m - 5m - 20\) | |
| A1 | for \(6m^3 + 11m^2 - 57m - 20\) |
Additional guidance
Note that, for example, \(3m - 20\) is regarded as three terms in the expansion of \((m + 4)(2m - 5)\)
First product must be a 3 or 4 term quadratic but need not be simplified or may be incorrect.
Accept \(a = 6\), \(b = 11\), \(c = -57\), \(d = -20\)