Higher November 2024 Paper 6 Q20
20 \((4x + a)(4x - a)(x^2 + 2) = 16x^4 + bx^2 - 50\)
Find the two possible pairs of values for \(a\) and \(b\).
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(a = 5\) \(b = 7\) \(a = -5\) \(b = 7\) with correct working | 6 | M1 for constant term identified as \(-2a^2\) or seen in their expansion M1 for their constant term \(= -50\) A1 for \(a = 5\) or \(a = -5\) | Correct working requires evidence of at least M1 AND M1 Full expansion is \(16x^4 + 4ax^3 - 4ax^3 + 32x^2 - a^2x^2 + 8ax - 8ax - 2a^2\) or better scores at least M1 AND M1 |
| AND M1 for \(b = 32 - a^2\) or for coefficient of \(x^2\) identified as \(32 - a^2\) or seen in their expansion M1 for \(b = 32 - (\textit{their } a)^2\) evaluated | May be seen as \(32x^2\) and \(-a^2x^2\) in the expansion Trials: M1M1M1M1 for \((4x + 5)(4x - 5)(x^2 + 2)\) expanded and simplified to \(16x^4 + 7x^2 - 50\) A1 for \(a = 5\) or \(a = -5\) or M1M1M1 for \((4x + 5)(4x - 5)(x^2 + 2)\) expanded to \(16x^4 + 20x^3 - 20x^3 + 32x^2 - 25x^2 + 40x - 40x - 50\) or better A1 for \(a = 5\) or \(a = -5\) | ||
| If 0, 1 or 2 scored instead award SC3 for both correct pairs of answers \(a = 5\) and \(b = 7\) \(a = -5\) and \(b = 7\) If 0 or 1 scored instead award SC2 for \(a = 5\) and \(a = -5\) or for \(a = 5\) and \(b = 7\) or for \(a = -5\) and \(b = 7\) If 0 scored instead award M1 for \((4x + a)(4x - a) = 16x^2 - a^2\) or for \((4x + a)(x^2 + 2) = 4x^3 + ax^2 + 8x + 2a\) or for \((4x - a)(x^2 + 2) = 4x^3 - ax^2 + 8x - 2a\) or SC1 for \(a = 5\) or \(a = -5\) or at least one answer with \(b = 7\) | |||