| (a) $x^3 + 3x^2 + 4x - 12 = 0 \Rightarrow x^3 + 3x^2 = 12 - 4x$ | M1 | Moves from $f(x)=0$, which may be implied by subsequent working, to $x^2(x+3) = \pm 12 \mp 4x$ by separating terms and factorising in either order. No need to factorise rhs |
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| $\Rightarrow x^2(x+3) = 12 - 4x \Rightarrow x^2 = \frac{12-4x}{(x+3)} \Rightarrow x = \sqrt{\frac{4(3-x)}{(x+3)}}$ | dM1A1* | Divides by $(x+3)$ term to make $x^2$ the subject, then takes square root. No need for rhs to be factorised. CSO. Do not allow sloppy algebra or notation with root on just numerator. The 12-4x needs to have been factorised. |
| (b) $x_1 = 1.41$, $awrt\ x_2 = 1.20$, $x_3 = 1.31$ | M1A1,A1 | An attempt to substitute $x_0 = 1$ into the iterative formula to calculate $x_1$. This can be awarded for the sight of $\sqrt{\frac{4(3-1)}{(3+1)}}, \sqrt{\frac{8}{4}}, \sqrt{2}$ and even 1.4. The subscript is not important. Mark as the first value found. $x_2 = awrt\ 1.20$, $x_3 = awrt\ 1.31$. Mark as the second and third values found. Condone 1.2 for $x_2$ |
| **(3)** | **(3)** | |
| (c) Choosing $(1.2715, 1.2725)$ or tighter containing root $1.271998323$ | M1 | |
| $f(1.2715) = (+)0.00827...$, $f(1.2725) = -0.00821...$ | M1 | Calculates $f(1.2715)$ and $f(1.2725)$, or the tighter interval with at least 1 correct to 1 sig fig rounded or truncated. Accept $f(1.2715) = -0.008$ 1st rounded or truncated $f(1.2715) = -0.01$ 2dp. Accept $f(1.2725) = (+)0.008$ 1sf rounded or truncated. Also accept $f(1.2725) = (+)0.01$ 2dp |
| Change of sign⟹$\alpha = 1.272$ | A1 | Both values correct. A valid reason; Accept change of sign, or >0 <0, or $f(1.2715) \times f(1.2725)<0$. And a (minimal) conclusion; Accept hence root or $\alpha = 1.272$ or QED or $\square$ |
| **(3)** | **(9 marks)** | |

**Alternative to (a) working backwards:**
| $x = \sqrt{\frac{4(3-x)}{(x+3)}} \Rightarrow x^2 = \frac{4(3-x)}{(x+3)} \Rightarrow x^2(x+3) = 4(3-x)$ | M1 | Square (both sides) and multiply by $(x+3)$ |
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| $x^3 + 3x^2 = 12 - 4x \Rightarrow x^3 + 3x^2 + 4x - 12 = 0$ | dM1 | Expand brackets and collect terms on one side of the equation = 0 |
| States that this is $f(x)=0$ | A1* | A statement to the effect that this is $f(x)=0$ |
| **(3)** | **(3)** | |

**Acceptable answer to (c) with an example of a tighter interval:**
| Choosing the interval $(1.2715, 1.2720)$. Contains root $1.2719(98323)$ | M1 |
|---|---|
| Calculates $f(1.2715)$ and $f(1.2720)$, or the tighter interval with at least 1 correct to 1 sig fig rounded or truncated. Accept $f(1.2715) = -0.008$ 1sf rounded or truncated $f(1.2715) = -0.01$ 2dp. Accept $f(1.2720) = (+)0.00003$ 1sf rounded or truncated $f(1.2720) = (+)0.00002$ 1sf rounded or truncated. Also accept $f(1.2720) = (+)0.00003$ 1sf rounded or truncated. Also accept $f(1.2720) = (+)0.00002$ 1sf rounded or truncated $f(1.2720) = (+)0.00003$ 1sf rounded or truncated | M1 |
| Both values correct (see above), A valid reason; Accept change of sign, or >0 <0, or $f(1.2715) \times f(1.2720)<0$. And a (minimal) conclusion; Accept hence root or $\alpha = 1.272$ or QED or $\square$ | A1 |

**Acceptable answer to (c) using $g(x)$ where $g(x) = \sqrt{\frac{4(3-x)}{(x+3)}} - x$:**
| 2nd M1 | Calculates $g(1.2715)$ and $g(1.2725)$, or the tighter interval with at least 1 correct to 1 sig fig rounded or truncated. $g(1.2715) = 0.0007559$. Accept $g(1.2715) = awrt\ (+)0.0008$ 1sf rounded or awrt 0.0007 truncated. $g(1.2725) = -0.0007610.5$. Accept $g(1.2725) = \ awrt\ -0.0008$ 1sf rounded or awrt -0.0007 truncated. |
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