| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One unknown constant: find it then solve |
| Difficulty | Moderate -0.8 This is a straightforward application of the Factor Theorem requiring substitution of x=-1 to find k, then factorising a cubic (which will have one linear factor given and a quadratic that factors nicely). It's routine C2 material with clear signposting and standard techniques, making it easier than average but not trivial since it requires multiple steps and factorisation. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| (a) \(f(-1) = 0 \therefore -1 - k - 20 = 0\) | M1 |
| \(k = -21\) | A1 |
| (b) Polynomial long division: | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((x + 1)(x^2 - x - 20) = 0\) | M1 | |
| \((x + 1)(x + 4)(x - 5) = 0\) | M1 | |
| \(x = -4, -1, 5\) | A1 | (6 marks) |
**(a)** $f(-1) = 0 \therefore -1 - k - 20 = 0$ | M1 |
$k = -21$ | A1 |
**(b)** Polynomial long division: | M1 A1 |
$x + 1 \mid x^3 + 0x^2 - 21x - 20$
- $x^3 + x^2$
- $-x^2 - 21x$
- $-x^2 - x$
- $-20x - 20$
- $-20x - 20$
$(x + 1)(x^2 - x - 20) = 0$ | M1 |
$(x + 1)(x + 4)(x - 5) = 0$ | M1 |
$x = -4, -1, 5$ | A1 | (6 marks)2.
$$f ( x ) = x ^ { 3 } + k x - 20 .$$
Given that $\mathrm { f } ( x )$ is exactly divisible by ( $x + 1$ ),
\begin{enumerate}[label=(\alph*)]
\item find the value of the constant $k$,
\item solve the equation $\mathrm { f } ( x ) = 0$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q2 [6]}}