| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Topic | Factor & Remainder Theorem |
| Type | Given factor, find all roots |
| Difficulty | Easy -1.3 Part (i) is trivial substitution requiring only arithmetic verification. Part (ii) is routine polynomial division to find a quadratic factor, then factorising that quadratic—standard algebraic manipulation with no problem-solving required. This is easier than average A-level content, being a straightforward early-paper question testing basic polynomial techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(x = 4\) into equation or attempt factorisation of \((x - 4)\) | M1 | |
| Verify \(y(4) = 0\) or that \((x - 4)\) is a factor | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^3 - 12x - 16 = (x - 4)(x^2 + 4x + 4)\) | B1 B1 | |
| \(= (x - 4)(x + 2)(x + 2)\) | B1 | [3] |
**Part (i)**
Substitute $x = 4$ into equation or attempt factorisation of $(x - 4)$ | M1
Verify $y(4) = 0$ or that $(x - 4)$ is a factor | A1 | [2]
**Part (ii)**
$x^3 - 12x - 16 = (x - 4)(x^2 + 4x + 4)$ | B1 B1
$= (x - 4)(x + 2)(x + 2)$ | B1 | [3]\begin{enumerate}[label=(\roman*)]
\item Show that $x = 4$ is a root of $x^3 - 12x - 16 = 0$. [2]
\item Hence completely factorise the expression $x^3 - 12x - 16$. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q1 [5]}}