| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two unknowns with show-that step |
| Difficulty | Moderate -0.3 This is a standard C2 factor/remainder theorem question requiring systematic application of the remainder theorem and factor theorem to find two unknowns. The steps are routine: substitute x=-2 for the remainder condition, substitute x=1/2 for the factor condition, solve simultaneous equations, then factorise. While multi-part, it follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(p(-2) = 20 \therefore -16 + 4 - 2a + b = 20\), so \(b = 2a + 32\) | M1, A1 | |
| (ii) \(p(\frac{1}{4}) = 0 \therefore \frac{1}{4} + \frac{1}{4} + \frac{1}{4}a + b = 0\). Substituting \(b = 2a + 32\): \(\frac{1}{2} + \frac{1}{4}a + (2a + 32) = 0\), so \(a = -13, b = 6\) | M1, M1, A2 | |
| (iii) Polynomial long division of \(2x^3 + x^2 - 13x + 6\) by \(2x - 1\) gives quotient \(x^2 + x - 6\). Therefore \(p(x) = (2x-1)(x^2 + x - 6) = (2x-1)(x+3)(x-2)\) | M1, A1 | (10) |
**(i)** $p(-2) = 20 \therefore -16 + 4 - 2a + b = 20$, so $b = 2a + 32$ | M1, A1 |
**(ii)** $p(\frac{1}{4}) = 0 \therefore \frac{1}{4} + \frac{1}{4} + \frac{1}{4}a + b = 0$. Substituting $b = 2a + 32$: $\frac{1}{2} + \frac{1}{4}a + (2a + 32) = 0$, so $a = -13, b = 6$ | M1, M1, A2 |
**(iii)** Polynomial long division of $2x^3 + x^2 - 13x + 6$ by $2x - 1$ gives quotient $x^2 + x - 6$. Therefore $p(x) = (2x-1)(x^2 + x - 6) = (2x-1)(x+3)(x-2)$ | M1, A1 | **(10)**
---8. The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 2 x ^ { 3 } + x ^ { 2 } + a x + b$$
where $a$ and $b$ are constants.\\
Given that when $\mathrm { p } ( x )$ is divided by $( x + 2 )$ there is a remainder of 20 ,\\
(i) find an expression for $b$ in terms of $a$.
Given also that $( 2 x - 1 )$ is a factor of $\mathrm { p } ( x )$,\\
(ii) find the values of $a$ and $b$,\\
(iii) fully factorise $\mathrm { p } ( x )$.\\
\hfill \mbox{\textit{OCR C2 Q8 [10]}}