| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.5 This is a straightforward C2 question testing basic understanding of the factor/remainder theorem. Part (a) requires simple substitution, part (b) applies the remainder theorem directly, and part (c) is routine factorisation once k is known. All steps are standard textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks |
|---|---|
| \(f(k) = -8\) | B1 |
| Answer | Marks |
|---|---|
| \(f(2) = 4 \Rightarrow 4 = (6 - 2)(2 - k) - 8\) | M1 |
| So \(k = -1\) | A1 |
| Answer | Marks |
|---|---|
| \(f(x) = 3x^2 - (2 + 3k)x + (2k - 8) = 3x^2 + x - 10\) | M1 |
| \(= (3x - 5)(x + 2)\) | M1A1 |
**Part (a)**
$f(k) = -8$ | B1
**Part (b)**
$f(2) = 4 \Rightarrow 4 = (6 - 2)(2 - k) - 8$ | M1 |
So $k = -1$ | A1 |
**Part (c)**
$f(x) = 3x^2 - (2 + 3k)x + (2k - 8) = 3x^2 + x - 10$ | M1 |
$= (3x - 5)(x + 2)$ | M1A1 |
**Guidance:**
**Part (b):** M1 for substituting $x = 2$ (not $x = -2$) and equating to 4 to form an equation in $k$. If the expression is expanded in this part, condone 'slips' for this M mark. Treat the omission of the $-8$ here as a 'slip' and allow the M mark.
**Beware:** Substituting $x = -2$ and equating to 0 (M0 A0) also gives $k = -1$.
**Alternative:** M1 for dividing by $(x - 2)$, to get $3x +$ (function of $k$), with remainder as a function of $k$, and equating the remainder to 4. [Should be $3x + (4 - 3k)$, remainder $-4k$].
**No working:** $k = -1$ with no working scores M0 A0.
**Part (c):**
$1^{\text{st}}$ M1 for multiplying out and substituting their (constant) value of $k$ (in either order). The multiplying-out may occur earlier. Condone, for example, sign slips, but if the 4 (from part (b)) is included in the $f(x)$ expression, this is M0. The $2^{\text{nd}}$ M1 is still available.
$2^{\text{nd}}$ M1 for an attempt to factorise their three term quadratic (3TQ).
A1 The correct answer, as a product of factors, is required.
Allow $3\left(x - \frac{5}{3}\right)(x + 2)$
Ignore following work (such as a solution to a quadratic equation).
If the 'equation' is solved but factors are never seen, the $2^{\text{nd}}$ M is not scored.
---3.
$$f ( x ) = ( 3 x - 2 ) ( x - k ) - 8$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $\mathrm { f } ( k )$.
When $\mathrm { f } ( x )$ is divided by $( x - 2 )$ the remainder is 4
\item Find the value of $k$.
\item Factorise f(x) completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2009 Q3 [6]}}