| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find remainder(s) then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the remainder theorem followed by routine factorisation. Part (a) requires simple substitution (f(2) and f(-3)), and part (b) uses the 'hence' to identify two linear factors, leaving a quadratic to factorise by inspection or formula. All steps are standard textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the cubic arithmetic involved. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(f(\pm2)\) or \(f(\pm3)\) | M1 | M1: As on scheme. Or uses long division as far as a remainder |
| (i) \(f(2) = 150\) | A1 | A1: for 150 |
| (ii) \(f(-3) = 0\) | A1 | A1: for 0. Both cao |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10x^3 + 27x^2 - 13x - 12 = (x+3)(10x^2 + \ldots)\) | M1 | M1: Recognises \((x+3)\) is factor and obtains correct first term of quadratic factor |
| \(10x^3 + 27x^2 - 13x - 12 = (x+3)(10x^2 - 3x - 4)\) | A1 | A1: Correct quadratic |
| \(``(10x^2 - 3x - 4)" = (ax+b)(cx+d)\) where \( | ac | =10\) and \( |
| \(= (x+3)(5x-4)(2x+1)\) | A1 | A1: Need all three factors. Accept e.g. \(10(x+3)(x-\frac{4}{5})(x+\frac{1}{2})\) |
| [4] | Special case: just writes three factors with no working – Full marks. Trial and error/calculator for completely correct answer – 4 marks or 0 if "hence" not used |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $f(\pm2)$ or $f(\pm3)$ | M1 | M1: As on scheme. Or uses long division as far as a remainder |
| (i) $f(2) = 150$ | A1 | A1: for 150 |
| (ii) $f(-3) = 0$ | A1 | A1: for 0. Both cao |
| | [3] | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $10x^3 + 27x^2 - 13x - 12 = (x+3)(10x^2 + \ldots)$ | M1 | M1: Recognises $(x+3)$ is factor and obtains correct first term of quadratic factor |
| $10x^3 + 27x^2 - 13x - 12 = (x+3)(10x^2 - 3x - 4)$ | A1 | A1: Correct quadratic |
| $``(10x^2 - 3x - 4)" = (ax+b)(cx+d)$ where $|ac|=10$ and $|bd|=4$ | dM1 | dM1: Attempt to factorise **their** quadratic |
| $= (x+3)(5x-4)(2x+1)$ | A1 | A1: Need all three factors. Accept e.g. $10(x+3)(x-\frac{4}{5})(x+\frac{1}{2})$ |
| | [4] | Special case: just writes three factors with no working – Full marks. Trial and error/calculator for completely correct answer – 4 marks or 0 if "hence" not used |
---3.
$$f ( x ) = 10 x ^ { 3 } + 27 x ^ { 2 } - 13 x - 12$$
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by
\begin{enumerate}[label=(\roman*)]
\item $x - 2$
\item $x + 3$
\end{enumerate}\item Hence factorise $\mathrm { f } ( x )$ completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q3 [7]}}