| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| 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 theorem question with routine steps: substitute x=2 to find the constant, perform polynomial division or comparison to factorise, then use remainder theorem. All techniques are direct applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(2)=-32+4a+18-18=0\) | M1 | Attempts \(f(2)\) or \(f(-2)\) |
| \(4a=32\Rightarrow a=8\) | A1 | cso |
| Way 2: | \(r=9\Rightarrow q=0\) also \(p=-4\therefore a=-2p=8\) | M1 A1 |
| Way 3: | \(4a-32=0\Rightarrow a=8\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x)=(x-2)(-4x^2+9)\) | M1 | Attempts long division or other method to obtain \((-4x^2\pm ax\pm b),\ b\neq0\), even with a remainder |
| \(=(x-2)(3-2x)(3+2x)\) or equivalent e.g. \(=-(x-2)(2x-3)(2x+3)\) or \(=(x-2)(2x-3)(-2x-3)\) | dM1 A1 | dM1: A valid attempt to factorise their quadratic — dependent on previous M1 being awarded, but there must have been no remainder. A1: cao — must have all 3 factors on the same line. Ignore subsequent work |
| Answer | Marks | Guidance |
|---|---|---|
| \(f\!\left(\frac{1}{2}\right)=-4\!\left(\frac{1}{8}\right)+8\!\left(\frac{1}{4}\right)+9\!\left(\frac{1}{2}\right)-18=-12\) | M1 A1ft | Attempts \(f\!\left(\frac{1}{2}\right)\) or \(f\!\left(-\frac{1}{2}\right)\); Allow A1ft for the correct numerical value of \(\frac{\text{their }a}{4}-14\) |
| Way 2: \(Q=-2x^2+3x+6,\ R=-12\) | M1 A1ft | M1: Attempt long division to give a remainder independent of \(x\); A1ft: correct numerical value of \(\frac{\text{their }a}{4}-14\) |
# Question 4(a):
| $f(2)=-32+4a+18-18=0$ | M1 | Attempts $f(2)$ or $f(-2)$ |
| $4a=32\Rightarrow a=8$ | A1 | cso |
**Way 2:** | $r=9\Rightarrow q=0$ also $p=-4\therefore a=-2p=8$ | M1 A1 | Compares coefficients leading to $-2p=a$; cso |
**Way 3:** | $4a-32=0\Rightarrow a=8$ | M1 A1 | Attempt to divide $\pm f(x)$ by $(x-2)$ to give quotient at least of the form $\pm4x^2+g(a)x$ and a remainder that is a function of $a$; cso |
**[2]**
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# Question 4(b):
| $f(x)=(x-2)(-4x^2+9)$ | M1 | Attempts long division or other method to obtain $(-4x^2\pm ax\pm b),\ b\neq0$, even with a remainder |
| $=(x-2)(3-2x)(3+2x)$ or equivalent e.g. $=-(x-2)(2x-3)(2x+3)$ or $=(x-2)(2x-3)(-2x-3)$ | dM1 A1 | dM1: A **valid** attempt to factorise their quadratic — dependent on previous M1 being awarded, but there must have been no remainder. A1: cao — must have all 3 factors on the same line. Ignore subsequent work |
**[3]**
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# Question 4(c):
| $f\!\left(\frac{1}{2}\right)=-4\!\left(\frac{1}{8}\right)+8\!\left(\frac{1}{4}\right)+9\!\left(\frac{1}{2}\right)-18=-12$ | M1 A1ft | Attempts $f\!\left(\frac{1}{2}\right)$ or $f\!\left(-\frac{1}{2}\right)$; Allow A1ft for the correct numerical value of $\frac{\text{their }a}{4}-14$ |
**Way 2:** $Q=-2x^2+3x+6,\ R=-12$ | M1 A1ft | M1: Attempt long division to give a remainder independent of $x$; A1ft: correct numerical value of $\frac{\text{their }a}{4}-14$ |
**[2] [Total 7]**4.\\
$\mathrm { f } ( x ) = - 4 x ^ { 3 } + a x ^ { 2 } + 9 x - 18$, where $a$ is a constant.
Given that ( $x - 2$ ) is a factor of $\mathrm { f } ( x )$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $a$,
\item factorise $\mathrm { f } ( x )$ completely,
\item find the remainder when $\mathrm { f } ( x )$ is divided by ( $2 x - 1$ ).
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q4 [7]}}