| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial with equal remainders |
| Difficulty | Standard +0.3 This is a standard C2 polynomial question testing the Remainder Theorem and factor theorem across multiple parts. Part (i) requires setting up two equations using the Remainder Theorem and solving simultaneously—straightforward algebra. Parts (ii)-(iii) are direct applications once k is found. Part (iv) requires factorizing to find one root, then analyzing a quadratic's discriminant, which is routine for C2. The question involves multiple steps but uses only standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| \(f(-1) = r \therefore -1 + k + 7 - 15 = r\) | M1 |
| \(k = r + 9\) | A1 |
| \(f(3) = 3r \therefore 27 + 9k - 21 - 15 = 3r\) | M1 |
| \(3k = r + 3\) | M1 |
| Subtracting, \(2k = -6\) | M1 |
| \(k = -3\) | A1 |
| Answer | Marks |
|---|---|
| \(r = -3 - 9 = -12\) | B1 |
| Answer | Marks |
|---|---|
| \(f(x) = x^3 - 3x^2 - 7x - 15\) | |
| \(f(5) = 125 - 75 - 35 - 15 = 0 \therefore (x - 5)\) is a factor | M1 A1 |
| Answer | Marks |
|---|---|
| \[\begin{array}{c | cc} |
| Answer | Marks | Guidance |
|---|---|---|
| \end{array}\] | M1 A1 | |
| \(\therefore (x - 5)(x^2 + 2x + 3) = 0\) | ||
| \(x = 5\) or \(x^2 + 2x + 3 = 0\) | ||
| \(b^2 - 4ac = 2^2 - (4 \times 1 \times 3) = -8\) | M1 | |
| \(b^2 - 4ac < 0 \therefore\) no real solutions to quadratic | A1 | |
| \(\therefore\) only one real solution | (12) |
| Answer | Marks |
|---|---|
| Total | (72) |
## Part (i)
$f(-1) = r \therefore -1 + k + 7 - 15 = r$ | M1 |
$k = r + 9$ | A1 |
$f(3) = 3r \therefore 27 + 9k - 21 - 15 = 3r$ | M1 |
$3k = r + 3$ | M1 |
Subtracting, $2k = -6$ | M1 |
$k = -3$ | A1 |
## Part (ii)
$r = -3 - 9 = -12$ | B1 |
## Part (iii)
$f(x) = x^3 - 3x^2 - 7x - 15$ | |
$f(5) = 125 - 75 - 35 - 15 = 0 \therefore (x - 5)$ is a factor | M1 A1 |
## Part (iv)
$$\begin{array}{c|cc}
& x^2 + 2x + 3 \\
x - 5 & x^3 - 3x^2 - 7x - 15 \\
& x^3 - 5x^2 \\
\hline
& 2x^2 - 7x \\
& 2x^2 - 10x \\
\hline
& 3x - 15 \\
& 3x - 15
\end{array}$$ | M1 A1 |
$\therefore (x - 5)(x^2 + 2x + 3) = 0$ | |
$x = 5$ or $x^2 + 2x + 3 = 0$ | |
$b^2 - 4ac = 2^2 - (4 \times 1 \times 3) = -8$ | M1 |
$b^2 - 4ac < 0 \therefore$ no real solutions to quadratic | A1 |
$\therefore$ only one real solution | | **(12)**
---
**Total** | **(72)** |The polynomial f(x) is given by
$$\text{f}(x) = x^3 + kx^2 - 7x - 15,$$
where $k$ is a constant.
When f(x) is divided by $(x + 1)$ the remainder is $r$.
When f(x) is divided by $(x - 3)$ the remainder is $3r$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $k$. [5]
\item Find the value of $r$. [1]
\item Show that $(x - 5)$ is a factor of f(x). [2]
\item Show that there is only one real solution to the equation f(x) = 0. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q9 [12]}}