| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Topic | Factor & Remainder Theorem |
| Type | Single unknown from factor condition |
| Difficulty | Moderate -0.3 Part (i) is a straightforward application of the factor theorem (substitute x=2, set equal to zero, solve for a). Part (ii) requires showing the quadratic factor has no real roots using the discriminant, which is standard technique but adds a small problem-solving element beyond pure recall. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
(i) Attempt to use factor theorem with $x = 2$ [M1]
Solve $40 + 4a + 12a - 8 = 0$, obtaining $a = -2$ [A1] **2 marks**
(ii) Factorise $5x^3 - 2x^2 - 12x - 8$ as $(x-2)(5x^2 + 8x + 4)$ [B1]
Attempt to find roots from quadratic or calculate its discriminant [M1]
State roots are complex and conclude that the cubic has only the one real root $x = 2$ [A1 (AG)] **3 marks**3 (i) Find the value of $a$ for which ( $x - 2$ ) is a factor of $5 x ^ { 3 } + a x ^ { 2 } + 6 a x - 8$.\\
(ii) Show that, for this value of $a$, the cubic equation $5 x ^ { 3 } + a x ^ { 2 } + 6 a x - 8 = 0$ has only one real root.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q3 [5]}}