# Question 1 (Factor Theorem):

## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts to calculate $f(\pm 1)$ | M1 | Algebraic division does not score this mark |
| Remainder is 2; $f(1) = 2$ | A1 | Accept sight of $f(1) = 2$ for both marks |

## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts to find $f(\pm 3)$; e.g. $-4\times(\pm3)^3 + 16\times(\pm3)^2 - 13\times(\pm3) + 3$ or $-108+144-39+3$ | M1 | Must see substitution or calculation; cannot accept just sight of '$f(3)=0$'; algebraic division does not score |
| $f(3) = -4\times(3)^3 + 16\times(3)^2 - 13\times(3) + 3 = 0 \Rightarrow (x-3)$ is a factor | A1* | "Show that" - requires statement AND conclusion (QED or tick); conclusion could be preamble before working |

## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Divides $f(x)$ by $(x-3)$ to get quadratic | M1 | If part (b) done by division, can be awarded here; inspection: look for first and last terms e.g. $f(x)=(x-3)(\pm4x^2+...x\pm1)$ |
| $(-4x^2 + 4x - 1)$ | A1 | |
| Further factorisation of $(-4x^2+4x-1)$ | dM1 | Dependent on previous M |
| $f(x)=(3-x)(2x-1)^2$, or $f(x)=-(x-3)(2x-1)^2$, or $f(x)=(x-3)(2x-1)(-2x+1)$ | A1 | $f(x)=4(3-x)\left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)$ also accepted; award M1A0M1A0 by implication for $(x-3)(2x-1)^2$; no marks for $(x-3)\left(x-\frac{1}{2}\right)^2$ |

## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sight of $\frac{1}{2}$ and/or $3$ | B1 | Seen in part (d) |
| Both $x=\frac{1}{2}$, $x\geq 3$ | B1 | Do not accept roots listed then $x\geq3$ added later; accept set language e.g. $\left\{x: x=\frac{1}{2} \cup x\geq3\right\}$ |

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