| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Prove root count with given polynomial |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing standard Factor and Remainder Theorem techniques with clear signposting. Parts (i) and (ii) are direct substitution, part (iii) is routine polynomial division/comparison, and part (iv) requires only discriminant checking or observation. No problem-solving insight needed, just methodical application of well-practiced procedures. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(-2) = -16 + 36 - 22 - 8 = -10\) | M1 | Attempt \(f(-2)\), or equiv |
| A1 (2) | Obtain \(-10\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(\frac{1}{2}) = \frac{1}{4} + 2\frac{1}{4} + 5\frac{1}{2} - 8 = 0\) AG | M1 | Attempt \(f(\frac{1}{2})\) (no other method allowed) |
| A1 (2) | Confirm \(f(\frac{1}{2}) = 0\), extra line of working required |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = (2x-1)(x^2 + 5x + 8)\) | M1 | Attempt complete division by \((2x-1)\) or \((x - \frac{1}{2})\) or equiv |
| A1 | Obtain \(x^2 + 5x + c\) or \(2x^2 + 10x + c\) | |
| A1 (3) | State \((2x-1)(x^2 + 5x + 8)\) or \((x - \frac{1}{2})(2x^2 + 10x + 16)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x)\) has one real root (\(x = \frac{1}{2}\)) because \(b^2 - 4ac = 25 - 32 = -7\) | B1\(\checkmark\) | State 1 root, following their quotient, ignore reason |
| hence quadratic has no real roots as \(-7 < 0\) | B1\(\checkmark\) (2) | Correct calculation, e.g. discriminant or quadratic formula, following their quotient, or cubic has max at \((-2.15, -9.9)\) |
## Question 7:
### Part (i):
| $f(-2) = -16 + 36 - 22 - 8 = -10$ | M1 | Attempt $f(-2)$, or equiv |
|---|---|---|
| | A1 **(2)** | Obtain $-10$ |
### Part (ii):
| $f(\frac{1}{2}) = \frac{1}{4} + 2\frac{1}{4} + 5\frac{1}{2} - 8 = 0$ AG | M1 | Attempt $f(\frac{1}{2})$ (no other method allowed) |
|---|---|---|
| | A1 **(2)** | Confirm $f(\frac{1}{2}) = 0$, extra line of working required |
### Part (iii):
| $f(x) = (2x-1)(x^2 + 5x + 8)$ | M1 | Attempt complete division by $(2x-1)$ or $(x - \frac{1}{2})$ or equiv |
|---|---|---|
| | A1 | Obtain $x^2 + 5x + c$ or $2x^2 + 10x + c$ |
| | A1 **(3)** | State $(2x-1)(x^2 + 5x + 8)$ or $(x - \frac{1}{2})(2x^2 + 10x + 16)$ |
### Part (iv):
| $f(x)$ has one real root ($x = \frac{1}{2}$) because $b^2 - 4ac = 25 - 32 = -7$ | B1$\checkmark$ | State 1 root, following their quotient, ignore reason |
|---|---|---|
| hence quadratic has no real roots as $-7 < 0$ | B1$\checkmark$ **(2)** | Correct calculation, e.g. discriminant or quadratic formula, following their quotient, or cubic has max at $(-2.15, -9.9)$ |
---7 The polynomial $\mathrm { f } ( x )$ is given by $\mathrm { f } ( x ) = 2 x ^ { 3 } + 9 x ^ { 2 } + 11 x - 8$.\\
(i) Find the remainder when $\mathrm { f } ( x )$ is divided by ( $x + 2$ ).\\
(ii) Use the factor theorem to show that ( $2 x - 1$ ) is a factor of $\mathrm { f } ( x )$.\\
(iii) Express $\mathrm { f } ( x )$ as a product of a linear factor and a quadratic factor.\\
(iv) State the number of real roots of the equation $\mathrm { f } ( x ) = 0$, giving a reason for your answer.
\hfill \mbox{\textit{OCR C2 2009 Q7 [9]}}