| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Known polynomial, verify then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring routine substitution to verify a given factor, followed by polynomial division and solving a quadratic. All steps are standard textbook procedures with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts \(f(\pm3)\) | M1 | Must use factor theorem |
| \(\{f(-3)=\}\ 0\) so \((x+3)\) is a factor of \(f(x)\) | A1 | For seeing 0 and conclusion (may be minimal e.g. QED, tick); need to see at least \((-3)^3+(-3)^2-12(-3)-18=0\) for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x^3+x^2-12x-18=(x+3)(x^2+\ldots)\) | M1 | Uses \((x+3)\) as factor, obtains correct first term of quadratic factor by division or comparing coefficients |
| \(x^3+x^2-12x-18=(x+3)(x^2-2x-6)\) or \((x+3)(x-1+\sqrt{7})(x-1-\sqrt{7})\) oe | A1 | Correct quadratic; this work may be done in part (a) and restated here |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((x=)-3\) | B1 | States \(-3\) |
| \(x=\frac{2\pm\sqrt{4+24}}{2}=1\pm\sqrt{7}\) or completion of square \((x-1)^2=7\) so \(x=1\pm\sqrt{7}\) or \((x-1+\sqrt{7})(x-1-\sqrt{7})=0\Rightarrow x=1\pm\sqrt{7}\) | M1 A1 | M1: Method for finding roots of their quadratic (for finding roots not just factors); check decimal answers (3.645…, −1.645…). A1: Both roots correct; allow \(x=\frac{2\pm\sqrt{28}}{2}\); extra roots e.g. \(x=-3,-1,\frac{2\pm\sqrt{28}}{2}\) loses final A mark |
## Question 6:
$f(x)=x^3+x^2-12x-18$
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $f(\pm3)$ | M1 | Must use factor theorem |
| $\{f(-3)=\}\ 0$ so $(x+3)$ is a factor of $f(x)$ | A1 | For seeing 0 and conclusion (may be minimal e.g. QED, tick); need to see at least $(-3)^3+(-3)^2-12(-3)-18=0$ for A1 |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $x^3+x^2-12x-18=(x+3)(x^2+\ldots)$ | M1 | Uses $(x+3)$ as factor, obtains correct first term of quadratic factor by division or comparing coefficients |
| $x^3+x^2-12x-18=(x+3)(x^2-2x-6)$ or $(x+3)(x-1+\sqrt{7})(x-1-\sqrt{7})$ oe | A1 | Correct quadratic; this work may be done in part (a) and restated here |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $(x=)-3$ | B1 | States $-3$ |
| $x=\frac{2\pm\sqrt{4+24}}{2}=1\pm\sqrt{7}$ or completion of square $(x-1)^2=7$ so $x=1\pm\sqrt{7}$ or $(x-1+\sqrt{7})(x-1-\sqrt{7})=0\Rightarrow x=1\pm\sqrt{7}$ | M1 A1 | M1: Method for finding roots of their quadratic (**for finding roots not just factors**); check decimal answers (3.645…, −1.645…). A1: Both roots correct; allow $x=\frac{2\pm\sqrt{28}}{2}$; extra roots e.g. $x=-3,-1,\frac{2\pm\sqrt{28}}{2}$ loses final A mark |6.
$$f ( x ) = x ^ { 3 } + x ^ { 2 } - 12 x - 18$$
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $( x + 3 )$ is a factor of $\mathrm { f } ( x )$.
\item Factorise $\mathrm { f } ( x )$.
\item Hence find exact values for all the solutions of the equation $\mathrm { f } ( x ) = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q6 [7]}}