| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Remainder condition then further work |
| Difficulty | Moderate -0.3 This is a standard C2 Factor/Remainder Theorem question with routine steps: substitute to find p, verify remainder using f(2), then factorise and solve a quadratic. All techniques are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| (i) \(-27 + 63 - 3p - 6 = 0\), \(p = 10\) | M1 A1 |
| (ii) remainder = \(f(2) = 8 + 28 + 20 - 6 = 50\) | M1 A1 |
| (iii) \(x = -3\) is a solution \(\therefore (x + 3)\) is a factor | B1 |
| \[x + 3 \mid \begin{array}{c} x^3 + 4x - 2 \\ x^3 + 7x^2 + 10x - 6 \\ x^3 + 3x^2 \\ \hline 4x^2 + 10x \\ 4x^2 + 12x \\ \hline -2x - 6 \\ -2x - 6 \end{array}\] | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| other solutions: \(x = \frac{-4 \pm \sqrt{16 + 8}}{2} = -4.45, 0.45\) | M1 A1 | (9) |
(i) $-27 + 63 - 3p - 6 = 0$, $p = 10$ | M1 A1 |
(ii) remainder = $f(2) = 8 + 28 + 20 - 6 = 50$ | M1 A1 |
(iii) $x = -3$ is a solution $\therefore (x + 3)$ is a factor | B1 |
$$x + 3 \mid \begin{array}{c} x^3 + 4x - 2 \\ x^3 + 7x^2 + 10x - 6 \\ x^3 + 3x^2 \\ \hline 4x^2 + 10x \\ 4x^2 + 12x \\ \hline -2x - 6 \\ -2x - 6 \end{array}$$ | M1 A1 |
$\therefore (x + 3)(x^2 + 4x - 2) = 0$
$x = -3$ or $x^2 + 4x - 2 = 0$
other solutions: $x = \frac{-4 \pm \sqrt{16 + 8}}{2} = -4.45, 0.45$ | M1 A1 | **(9)** |
---6. Given that
$$f ( x ) = x ^ { 3 } + 7 x ^ { 2 } + p x - 6$$
and that $x = - 3$ is a solution to the equation $\mathrm { f } ( x ) = 0$,\\
(i) find the value of the constant $p$,\\
(ii) show that when $\mathrm { f } ( x )$ is divided by $( x - 2 )$ there is a remainder of 50 ,\\
(iii) find the other solutions to the equation $\mathrm { f } ( x ) = 0$, giving your answers to 2 decimal places.\\
\hfill \mbox{\textit{OCR C2 Q6 [9]}}